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4 Qualitative theory of differential equations

4.1 Introduction In this chapter we consider the differential equation

x=f(t,x)

where

and

f(t,x)=

fn(t,Xp ... ,xn)

(1)

is a nonlinear function of x1, ... , xn. Unfortunately, there are no known methods of solving Equation (1). This, of course, is very disappointing. However, it is not necessary, in most applications, to find the solutions of (1) explicitly. For example, let x 1(t) and xit) denote the populations, at time t, of two species competing amongst themselves for the limited food and living space in their microcosm. Suppose, moreover, that the rates of growth of x1(t) and xit) are governed by the differential equation (1). In this case, we arenot really interested in the va1ues of x 1(t) and x2(t) at every time t. Rather, we are interested in the qualitative properties of x1(t) and x2(t). Specically, we wish to answer the following questions.

372

4.1 lntroduction

I. Do there exist values ~ 1 and ~2 at which the two species coexist together in a steady state? That is to say, are there numbers ~ 1 , ~2 such that x 1 (t)=~1 , xit)=~2 is a solution of (1)? Such values ~ 1 ,~2, if they exist, are called equilibrium points of (I).

2. Suppose that the two species are coexisting in equilibrium. Suddenly, we add a few members of species 1 to the microcosm. Will x 1(t) and x2(t) remain close to their equilibrium values for all future time? Or perhaps the extra few members give species I a large advantage and it will proceed to annihilate species 2.

3. Suppose that x 1 and x2 have arbitrary values at t=O. What happens as t approaches infinity? Will one species ultimately emerge victorious, or will the struggle for existence end in a draw?

More generally, we are interested in determining the following properties of solutions of (1).

1. Do there exist equilibrium values

for which x(t)=::x0 is a solution of (I)? 2. Let <f> (t) be a solution of (I). Suppose that 1/;(t) is a second solution

with 1/;(0) very close to <f> (0); that is, 1/;;(0) is very close to <MO), j = I, ... , n. Will 1/;(t) remain close to <f> ( t) for all future time, or will 1/;( t) diverge from </> (t) as t approaches infinity? This question is often referred to as the problern of stability. lt is the most fundamental problern in the qualitative theory of differential equations, and has occupied the attention of many mathematicians for the past hundred years.

3. What happens to solutions x(t) of (I) as t approaches infinity? Do all solutions approach equilibrium values? If they don't approach equilibrium values, do they at least approach a periodic solution?

This chapter is devoted to answering these three questions. Remarkably, we can often give satisfactory answers to these questions, even though we cannot solve Equation (I) explicitly. Indeed, the first question can be answered immediately. Observe that x(t) is identically zero if x(t) = x0•

Hence, x0 is an equilibrium value of (1), if, and only if,

(2)

Example 1. Find all equilibrium values of the system of differential equations

373


Solution.

is an equilibrium value if, and only if, 1-x~=O and (x?)3 +x~=O. This

implies that x~ = 1 and x? = - 1. Hence ( - ~ ) is the only equilibrium value

of this system.

Example 2. Find all equilibrium solutions of the system

~~ =(x-l)(y-1), dy dt = (X + } )(y + 1 ).

Solution.

is an equilibrium value of this system if, and only if, (x0 -1)(y0 -1)=0 and (x0 + l)(y0 + 1)=0. The first equation is satisfied if either x 0 or y 0 is 1, while the second equation is satisfied if either x0 or y 0 is - 1. Hence, x = 1, y = - I and x = - I, y = I are the equilibrium solutions of this system.

The question of stability is of paramount importance in all physica1 applications, since we can never measure initial conditions exactly. For example, consider the case of a particle of mass one kgm attached to an elastic spring of force constant 1 Njm which is moving in a frictionless medium. In addition, an external force F(t) = cos2t N is acting on the particle. Let y( t) denote the position of the particle relative to its equilibrium position. Then (d 2yjdt 2 )+ y = cos2t. We convert this second-order equation into a system of two first-order equations by setting x 1 = y, x 2 = y'. Then,

dx2 ---;[( =- x 1 + cos2t. (3)

The functionsy 1(t)=sint andyit)=cost are two independent solutions of the homogeneous equationy"+y=O. Moreover,y= -icos2t is a particular solution of the nonhomogeneous equation. Therefore, every solution

of (3) is of the form

x(t)=cl(sint)+cz( cC?st)+[-icos2t ]· cost -smt jsin2t

(4)

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4.1 Introduction

At time t = 0 we measure the position and velocity of the particle and obtainy(O)= l,y'(O)=O. This implies that c1 =0 and c2= j. Consequently, the position and velocity of the particle for all future time are given by the equation

( y(t)) (x1(t)) [ jcost-tcos2tl y'(t) = x 2(t) = - jsint+ ~sin2t ·

(5)

However, suppose that our measurements permit an error of magnitude 10-4• Will the position and velocity of the particle remain close to the values predicted by (5)? The answer to this question had better be yes, for otherwise, Newtonian mechanics would be of no practical value to us. Fortunately, it is quite easy to show, in this case, that the position and velocity of the particle remain very close to the values predicted by (5). Lety(t) and j'(t) denote the true values of y(t) and y'(t) respectively. Clearly,

y(t)-y (t) = (j- c2)cost- c1 sint

y'(t)- .Y'(t) =- c1 cost-{j- c2)sint

where c1 and c2 are two constants satisfying

We can rewrite these equations in the form

[ 2]1/2 y'(t)-.Y'(t)= cr+O-c2) cos(t-«52),

c, tan«51=--4

c2-3

~-c 3 2 tan«52=--.

c,

Hence, both y ( t)-y ( t) and y' ( t)-y' ( t) are bounded in absolute value by [cf +(j- c2)2]112. This quantity is at most V2 10-4• Therefore, the true values of y(t) and y'(t) are indeed close to the values predicted by (5).

As a second example of the concept of stability, consider the case of a particle of mass m which is supported by a wire, or inelastic string, of length I and of negligible mass. The wire is always straight, and the system is free to vibrate in a vertical plane. This configuration is usually referred to as a simple pendulum. The equation of motion of the pendulum is

where y is the angle which the wire makes with the vertical line AO (see

375


Figure 1

Figure 1). Setting x 1 = y and x2 = dy / dt we see that

(6)

The system of equations (6) has equilibrium solutions x 1 =0, x2 =0, and x 1

= '17, x 2 = 0. (If the pendulum is suspended in the upright position y = '1T

with zero velocity, then it will remain in this upright position for all future time.) These two equilibrium solutions have very different properties. If we disturb the pendulum slightly from the equilibrium position x 1 = 0, x2 = 0, by either displacing it slightly, or giving it a small velocity, then it will execute small oscillations about x 1 =0. On the other band, if we disturb the pendulum slightly from the equilibrium position x 1 = '17, x2 =0, then it will either execute very large oscillations about x 1 =0, or it will rotate around and around ad infinitum. Thus, the slightest disturbance causes the pendulum to deviate drastically from its equilibrium position x 1 = '17, x 2 = 0. Intuitively, we would say that the equilibrium value x 1 =0, x2 =0 of (6) is stable, while the equilibrium value x 1 = '17, x 2 = 0 of (6) is unstable. This concept will be made precise in Section 4.2.

The question of stability is usually very difficult to resolve, because we cannot solve (1) explicitly. The only case which is manageable is when f(t, x) does not depend explicitly on t; that is, f is a function of x alone. Such differential equations are called autonomous. And even for autonomaus differential equations, there are only two instances, generally, where we can completely resolve the stability question. The first case is when f(x) = Ax and it will be treated in the next section. The second case is when we are only interested in the stability of an equilibrium solution of x=f(x). This case will be treated in Section 4.3.

Question 3 is extremely important in many applications since an answer to this question is a prediction concerning the long time evolution of the system under consideration. We answer this question, when possible, in Sections 4.6-4.8 and apply our results to some extremely important applications in Sections 4.9-4.12.

376

4.1 Introduction

EXERCISES

In each of Problems 1-8, find all equilibrium values of the given system of differential equations.

1. ~~ =x-x2-2xy

dy =2y-2y2-3xy dt

dx 3. dt = ax- bxy

~ = -cy+dxy

dz =z+x2+y2 dt

5. dx =xy2-x dt dy 0

dt =XSID'TT)'

7. ~~ = - 1-y- ex

dy =x2+y(ex-1) dt dz 0

dt =x+smz

dx 2. dt=-ßxy+JL

~ =ßxy-yy

dx 2 4. dt= -x-xy

dy 2 dt = -y-yx

dz = 1-z+x2 dt

dx 6. dt =cosy

dy 0 1 dt =smx-

8. ~~ =x-y2

dy =x2-Y dt

dz =ez-x dt

9. Consider the systern of differential equations

dx dy dt =ax+by, dt =cx+dy.

(i) Show that x=O,y=O is the only equilibriurn point of (*) if ad-bc=/=0. (ii) Show that (*) has a line of equilibriurn points if ad- bc = 0.

10. Let x = x(t), y = y(t) be the solution of the initial-value problern

dx dt =-x-y,

dy dt =2x-y, x(O)= y(O) = 1.

(*)

Suppose that we rnake an error of rnagnitude w-4 in rneasuring x(O) andy(O). What is the largest error we rnake in evaluating x(t), y(t) for 0.;;;; t < oo?

11. (a) Verify that

is the solution of the initial-value problern

x=( i -3

l 1 2

x(O)= (g). 377


(b) Let x=I/J(t) be the solution of the above differential equation which satisfies the initial condition

x(O)=x0~U). Show that each component of 1/J(t) approaches infinity, in absolute value, as t-Hx;.

4.2 Stability of linear systems

In this section we consider the stability question for solutions of autonomaus differential equations. Let x=f[>(t) be a solution of the differential equation

x=f(x). (I)

We are interested in determining whether f[>(t) is stable or unstable. That is to say, we seek to determine whether every solution l[;(t) of (l) which starts sufficiently close to 4>( t) at t = 0 must remain close to f[>( t) for all future time t ~ 0. We begin with the following formal definition of stability.

Definition. The solution x=f[>(t) of (1) is stable if every solution l[;(t) of (1) which starts sufficiently close to 4>( t) at t = 0 must remain close to 4>( t) for all future time t. The solution f[>(t) is unstable if there exists at least one solution l[;(t) of (I) which starts near f[>(t) at t=O but which does not remain close to f[>(t) for all future time. More precisely, the solution f[>(t) is stable if for every e > 0 there exists 8 = 8 ( e) such that

ilf1(t)-cpj(t)i<e if ll/li0)-!f>;(O)I<8(e), j=I, ... ,n

for every solution l[;(t) of (I).

The stability question can be completely resolved for each solution of the linear differential equation

x=Ax. (2)

This is not surprising, of course, since we can solve Equation (2) exactly. Wehave the following important theorem.

Theorem 1. (a) Every solution x=f[>(t) of (2) is stable if all the eigenvalues of A have negative real part. (b) Every solution x = f[>( t) of (2) is unstable if at least one eigenvalue of A has positive real part. (c) Suppose that all the eigenvalues of A have real part ...; 0 and A1 = io1, ••• ,A1 = io1 have zero real part. Let Aj = ioj have multip/icity kj. This means that the characteristic po/ynomial of A can be factared into the form

p(A) = (A- io1)k' . .. (A- io1)k'q(A)

378


where all the roots of q(A.) have negative real part. Then, every solution x=f/>(t) of (1) is stable if A has ky linearly independent eigenvectors for each eigenvalue A.1 = ia1. Otherwise, every solution f/>(t) is unstable.

Our first step in proving Theorem 1 is to show that every solution f/>(t) is stable if the equilibrium solution x(t)=O is stable, and every solution f/>(t) is unstable if x(t)=O is unstable. To this end, Iet t/;(t) be any solution of (2). Observe that z( t) = f/>( t)- t/;( t) is again a solution of (2). Therefore, if the equilibrium solution x(t)=O is stable, then z(t) =</>( t)- 1/;(t) will always remain small if z(O) =f/>(0)- t/;(0) is sufficiently small. Consequently, every solution f/>(t) of (2) is stable. On the other hand suppose that x(t)=O is unstable. Then, there exists a solution x = h(t) which is very small initially, but which becomes !arge as t approaches infinity. The function 1/;(t) =f/>(t) +h(t) is clearly a solution of (2). Moreover, t/;(t) is close to f/>(t) initially, but diverges from f/>(t) as t increases. Therefore, every solution x=f/>(t) of (2) is unstable.

Our next step in proving Theorem I is to reduce the problern of showing that n quantities 1/;/ t), j = 1, ... , n are small to the much simpler problern of showing that only one quantity is small. This is accomplished by introducing the concept of length, or magnitude, of a vector.

Definition. Let

be a vector with n components. The numbers x 1, ••• ,xn may be real or complex. We define the length of x, denoted by llxl/ as

1/xll =max{lxii, lx2l• · · ., lxnl}·

For example, if

then llxll = 3 and if

[ 1 + 2il

x= 2 -1

then llxl/ = V5 . The concept of the length, or magnitude of a vector corresponds to the

concept of the length, or magnitude of a number. Observe that llxll ~ 0 for

379


any vector x and llxll = 0 only if x = 0. Second, observe that

IIAxll =max{IXxli, ... ,IXxnl} =IXImax{ixii, ... , lxnl} =IXIIIxll·

Finally, observe that

llx+yll =max{lxi +yd, ... ,ixn +Yni} ~max{lxii +IYII, ... ,Ixnl +IYnl}

~ max{lxii, ... , lxnl} +max{iyd, ... ,IYnl} = llxll + IIYII·

Thus, our definition really captures the meaning of length. In Section 4.7 we give a simple geometric proof of Theorem 1 for the

case n = 2. The following proof is valid for arbitrary n.

PRooF OF THEOREM 1. (a) Every solution x=l[;(t) of i=Ax is of the form l[;(t) = eA1l[;(O). Let <Pv(t) be the ij element of the matrix eA1, and Iet 1/1?, ... , 1/12 be the components of 1[;(0). Then, the ith component of 1[;( t) is

n

1/1;( t) =<Pi! ( t)t/1? + ... + <P;n( t)l/J2= ~ <~>v( t)l/Jr j=l

Suppose that all the eigenvalues of A have negative real part. Let - a 1 be the largest of the real parts of the eigenvalues of A. It is a simple matter to show (see Exercise 17) that for every number - a, with - a 1 < - a < 0, we can find a number K such that I<Pv(t)l ~ Ke-a1, t ~ 0. Consequently,

n n

11/J;(t)l ~ ~ Ke-atll/J}I = Ke-at ~ ll/Jj01 j=l j=l

for some positive constants K and a. Now, 11[;1°1 ~ 111[;(0)11· Hence,

II 1/;( t) II = max{ ll/J1 ( t)l, · · ·, 11/Jn(t)l} ~ nKe -arll 1[;(0)11·

Let e>O be given. Choose 8(e)=ejnK. Then, 111/;(t)ll <e if 111[;(0)11 <8(e) and t ;;;. 0, since

111/;(t)ll ~ nKe-a1 lll[;(O)II < nKe/ nK = e.

Consequently, the equilibrium solution x(t)=O is stable. (b) Let X be an eigenvalue of A with positive real part and let v be an

eigenvector of A with eigenvalue X. Then, 1[;( t) = ce>-1v is a solution of i = Ax for any constant c. If Ais real then v is also real and 111/;(t)ll = lcle>-1 llvll· Clearly, 111/;(t)ll approaches infinity as t approaches infinity, for any choice of c~O, no matter how small. Therefore, x(t)=:O is unstable. If X= a + iß is complex, then v=v1 + iv2 is also complex. In this case

e<a+iß)l(v1 + ir) = eal (cosßt + isinßt)(v1 + iv2)

= ea1[ (v1 cosßt -rsinßt) + i(v1 sinßt+v2 cosßt)]

is a complex-valued solution of (2). Therefore

1[; 1 (t) = cear (v1 cosßt -rsinßt)

is a real-valued solution of (2), for any choice of constant c. Clearly,

380


!llf1(t)11 is unbounded as t approaches infinity if c and either v1 or v2 is nonzero. Thus, x(t)=O is unstable.

(c) If A has kj linearly independent eigenvectors for each eigenvalue 'J.y= iaj of multiplicity kj> then we can find a constant K suchthat l(eA1)1il < K

(see Exercise 18). There, lllf(t)ll < nKII~f(O)II for every solution lf(t) of (2). lt now follows immediately from the proof of (a) that x(t):=O is stable.

On the other hand, if A has fewer than Js linearly independent eigenvectors with eigenvalue 'J.y = iaj> then :X= Ax has solutions lf( t) of the form

lf( t) = ce;"i1 [ v + t(A- ia})v]

where (A-i~l)v~O. If ~=0, then lf(t)=c(v+tAv) is real-valued. Moreover, !llf(t)ll is unbounded as t approaches infinity for any choice of c~O. Similarly, both the real and imaginary parts of lf(t) are unbounded in magnitude for arbitrarily small lf(O) ~ 0, if aj ~ 0. Therefore, the equilibrium solution x(t):=O is unstable. 0

If all the eigenvalues of A have negative real part, then every solution x(t) of x=Ax approaches zero as t approaches infinity. This follows immediately from the estimate llx(t)ll < Ke-"1 llx(O)II which we derived in the proof of part (a) of Theorem I. Thus, not only is the equilibrium solution x(t)=O stable, but every solution lf(t) of (2) approaches it as t approaches infinity. This very strong type of stability is known as asymptotic stability.

Definition. A solution x=«f>(t) of (I) is asymptotically stableifit is stable, and if every solution lf( t) which starts sufficiently close to «/>( t) must approach «f>(t) as t approaches infinity. In particular, an equilibrium solution x(t) = x0 of (I) is asymptotically stable if every solution x = lf( t) of (I) which starts sufficiently close to x0 at time t = 0 not only remains close to x0 for all future time, but ultimately approaches x0 as t approaches infinity.

Remark. The asymptotic stability of any solution x=«f>(t) of (2) is clearly equivalent to the asymptotic stability of the equilibrium solution x(t)=O.

Example 1. Determine whether each solution x(t) of the differential equation

[-1

x= -2 -3

-~ ~]x -2 -1

is stable, asymptotically stable, or unstable. Solution. The characteristic polynomial of the matrix

A=[=~ -~ ~] -3 -2 -1

381


is

p(A)=(detA-AI)=det[-~;A -1°-A ~ l -3 -2 -1-A

= -(1 +A)3 -4(1 +A)= -(1 +A)(A2 +2A+5).

Hence, A = - 1 and A = - 1 ± 2i are the eigenvalues of A. Since all three eigenvalues have negative real part, we conclude that every solution of the differential equation x = Ax is asymptotically stable.

Example 2. Prove that every solution of the differential equation

is unstable. Solution. The characteristic po1ynomial of the matrix

is

Hence A = 6 and A = - 4 are the eigenvalues of A. Since one eigenvalue of A is positive, we conclude that every solution x = f/'( t) of x = Ax is unstable.

Example 3. Show that every solution of the differential equation

x=(o -3) 2 0 X

is stable, but not asymptotically stable. Solution. The characteristic polynomia1 of the matrix

is

p(A)=det(A-AI)=det( -; =n=A2 +6.

Thus, the eigenvalues of Aare A= ± V6 i. Therefore, by part (c) of Theorem 1, every solution x=f/'(t) of x=Ax is stable. However, no solution is asymptotically stab1e. This follows immediate1y from the fact that the general solution of x = Ax is

x(t)=c1(-V6 sinv'6t)+c2 (V6 cosv'6t). 2cos V6 t 2sin V6 t

382


Hence, every solution x(t) is periodic, with period 2'1T jv'6, and no solution x(t) (except x(t)=:O) approaches 0 as t approaches infinity.

Example 4. Show that every solution of the differential equation

x= [ ~ -6

-3 -6

0 -~]x -3

is unstable. Solution. The characteristic polynomial of the matrix

is

A= [ ~ -6

[2-X

p(X)=det(A-XI)=det 0 -6

-3 -6

0 -~] -3

-3 -6-X

0

Hence, the eigenvalues of A are X= - 7 and X= 0. Every eigenvector v of A with eigenvalue 0 must satisfy the equation

Av=[ ~ -6

-3 -6

0

This implies that v1 = 3v2/2 and v3 = - 3v2, so that every eigenvector v of A with eigenvalue 0 must be of the form

Consequently, every solution x=f[>(t) of x=Ax is unstable, since X=O is an eigenvalue of multiplicity two and A has only one linearly independent eigenvector with eigenvalue 0.

EXERCISES

Determine the stability or instability of all solutions of the following systems of differential equations.

1. x={ _J _J)x . ( -3 2. x= 2 -4) I X

3. x=( -s -I nx 4. x=(! -4) -7 X

( -7 I -6) 6. i=U

2 nx 5. i= I~ -4 Ii X 0 -I 2

383


1. x=( -1 2 -qx

( -2 I ~)x -3 s. x= -i 2 I -I -I -2

[ 0 2 0

~]x 10. ·~ [ -~ 2 I

~]x . -2 0 0 0 0 9. x= g 0 0 0 0

0 -2 0 -2

11. Determine whether the solutions x(t)=:=O and x(t)= I of the single scalar equation x = x(I- x) are stable or unstable.

12. Determine whether the solutions x(t)=:=O and x(t)= I of the single scalar equation x = - x(I- x) are stable or unstable.

13. Consider the differential equation x = x 2• Show that all solutions x(t) with x(O) ;;. 0 are unstable while all solutions x(t) with x(O) < 0 are asymptotically stable.

14. Consider the system of differential equations

(*)

(a) Show that

x(t)=(x1(t))=(csin(ct+d) ) x 2 (t) c2 cos(ct+d)

is a solution of (*) for any choice of constants c and d. (b) Assurne that a solution x(t) of (*) is uniquely determined once x 1(0) and

x2(0) are prescribed. Prove that (a) represents the general solution of (*). (c) Show that the solution x=O of (*) is stable, but not asymptotically stable. ( d) Show that every solution x( t) =ji 0 of (*) is unstable.

15. Show that the stability of any solution x( t) of the nonhomogeneaus equation x = Ax + f(t) is equivalent to the stability of the equilibrium solution x = 0 of the homogeneous equation x = Ax.

16. Determine the stability or instability of all solutions x(t) of the differential equation

. (-I x= 2

17. (a) Let f(t) = t 0e -bt, for some positive constants a and b, and Iet c be a positive number smaller than b. Show that we can find a positive constant K such that Jf(t)J< Ke-c1, 0 < t < oo. Hint: Show that f(t)/ e-ct approaches zero as t approaches infinity.

(b) Suppose that all the eigenvalues of A have negative real part. Show that we can find positive constants K and a suchthat J(eA1)iiJ .;;; Ke-a1 for I< i, j.;;; n. Hint: Each component of eA1 is a finite linear combination of functions of the form q(t)e111, where q(t) is a polynomial in t (of degree .;;; n -1) and A is an eigenvalue of A.

384

4.3 Stability of equilibrium solutions

18. (a) Let x(t)= e 1" 1v, o real, be a complex-valued solution of x=Ax. Show that both the real and imaginary parts of x( t) are bounded solutions of x = Ax.

(b) Suppose that all the eigenvalues of A have real part .;;; 0 and X1 = io1, ... , ~ = io1 have zero real part. Let X1 = io1 have multiplicity ki' and suppose that A has k1 linearly independent eigenvectors for each eigenvalue X1,} = I, ... ,1. Prove that we can find a constant K suchthat l(eA1)!il< K.

19. Let

and define llxlh = lxd + ... + lxnl· Show that (i) llxiii~Oand llxll 1=0onlyifx=O

(ii) IIXxll, = IXIIIxll, (iii) llx + Yll1 < llxiii + IIYIII

20. Let

and define llxll2 =[lxd2 + ... + lxnl2] 112• Show that (i) llxll2 ~ 0 and llxllz = 0 only if x = 0

(ii) IIXxllz = IXIIIxllz (iii) llx + Yllz < llxllz + IIYIIz

21. Show that there exist constants M and N such that

Mllxiii < IIXIIz < Nllxiii·


In Section 4.2 we treated the simple equation x = Ax. The next simplest equation is

x=Ax+g(x) (1)

where

g(x)=

is very small compared to x. Specifically we assume that

g1 (x) gn(x)

max{lx1l, ... ,lxnl} , ... , max{lx1l, ... ,lxnl}

385


are continuous functions of x 1, ... , xn which vanish for x 1 = ... = xn = 0. This is always the case if each component of g(x) is a polynomial in x 1, ••• ,xn which begins with terms of order 2 or higher. For example, if

( x x2) g(x)= 1 2 , xlx2

then both x1xi/max{lx11, lx21} and x 1x2jmax{lx11, lx21} are continuous functions of x 1,x2 which vanish for x 1 =x2=0.

If g(O)=O then x(t)=O is an equilibrium solution of (1). We would like to determine whether it is stable or unstable. At first glance this would seem impossible to do, since we cannot solve Equation (I) explicitly. However, if x is very small, then g(x) is very small compared to Ax. Therefore, it seems plausible that the stability of the equilibrium solution x(t)=O of (1) should be determined by the stability of the "approximate" equation x = Ax. This is almost the case as the following theorem indicates.

Theorem 2. Suppose that the vector-valued function

g(x)/llxll =g(x)jmax{ lx1!, · · ·, lxnl}

is a continuous function of x 1, ... ,xn which vanishes for x=O. Then, (a) The equilibrium solution x(t):=O oj (I) is asymptotically stable if the equilibrium solution x(t)=O oj the "linearized" equation x=Ax is asymptotically stable. Equivalently, the solution x(t)=O of (1) is asymptotically stable if all the eigenvalues of A have negative real part. (b) The equilibrium solution x(t)=O of (1) is unstable if at least one eigenvatue of A has positive real part. (c) The stability of the equilibrium solution x(t)=O of (l) cannot be determined from the stability oj the equilibrium solution x( t) = 0 of x = Ax if all the eigenvalues oj A have real part ~ 0 but at least one eigenvalue of A has zero real part.

PROOF. (a) The key step in many stability proofs is to use the variation of parameters formula of Section 3.12. This formu1a implies that any solution x(t) of (1) can be written in the form

(2)

We wish to show that llx(t)il approaches zero as t approaches infinity. To this end recall that if all the eigenvalues of A have negative real part, then we can find positive constants K and a such that (see Exercise 17, Section 4.2).

and

lleA(t-•)g(x(s))ll ~ Ke-a(t-s)llg(x(s))ll·

386


Moreover, we can find a positive constant a such that

llg(x)ll < 2~ llxll if llxll < a.

This follows immediately from our assumption that g(x)/llxll is continuous and vanishes at x = 0. Consequently, Equation (2) implies that

llx(t)ll < lleA1x(O)II + fo11leA(t-s)g(x(s))ll ds

< Ke-a1llx(O)II +!!:.. r1e-a(t-s)llx(s)llds 2 Jo

as long as llx(s)ll < o, 0 < s < t. Multiplying both sides of this inequality by eat gives

(3)

The inequality (3) can be simplified by setting z(t)=ea1llx(t)ll, for then

a (t z(t) < Kllx(O)II + 2 Jo z(s)ds. (4)

We would like to differentiate both sides of (4) with respect to t. However, we cannot, in general, differentiate both sides of an inequality and still preserve the sense of the inequality. We circumvent this difficulty by the clever trick of setting

Then

or

a ( U(t)= 2 Jo z(s)ds.

dU(t) a aK ----;[(- 2 U(t) < Tllx(O)II·

Multiplying both sides of this inequality by the integrating factor e-at/2

gives

or

~e-at/2 [ U(t)+KIIx(O)II] <0.

Consequently,

e-at/2 [ U(t) + Kllx(O)II] < U(O) + Kllx(O)II = Kllx(O)II,

387


so that U(t)< -KJJx(O)JJ+KIIx(O)JJeat/2• Returning to the inequality (4), we see that

JJx(t)/1 = e-atz(t) < e-at[ Kllx(O)JI + U (t)]

.;;;; Kllx(O)IIe-at/2 (5)

as long as llx(s)J/ < a, 0 < s < t. Now, if llx(O)II < a I K, then the inequality (5) guarantees that llx(t)ll < a for all future time t. Consequently, the inequality (5) is true for all t;;;. 0 if llx(O)II < a I K. Finally, observe from (5) that llx(t)ll < Kllx(O)II and llx(t)ll approaches zero as t approaches infinity. Therefore, the equilibrium solution x(t)=O of (1) is asymptotically stable. (b) The proof of (b) is too difficult to present here. (c) We will present two differential equations of the form (I) where the nonlinear term g(x) determines the stability of the equilibrium solution x(t) :=0. Consider first the system of differential equations

dxl- 2 2 dt -x2-xl(xl +x2), (6)

The linearized equation is

and the eigenvalues of the matrix

(- ~ ~) are ± i. To analyze the behavior of the nonlinear system (6) we multiply the first equation by x 1, the second equation by x2 and add; this gives

dxl dx2 _ 2( 2 2) 2( 2 2) xl dt +x2 dt - -xl xl +x2 -x2 xl +x2

But

Hence,

This implies that

where

388

= -(xi+x~(

xf(t)+x~(t)= -1 c2 ' + ct

c = xf(O) + x~(O).


Thus, xf( t) + x~( t) approaches zero as t approaches infinity for any solution x 1(t), xit) of (6). Moreover, the value of xf+x~ at any timet is always less than its value at t=O. We conclude, therefore, that x 1(t)=O, xit) =0 is asymptotically stable.

On the other band, consider the system of equations

dxl 2 2 dt =x2+xi(xi +x2),

Here too, the linearized system is

x={ _ ~ ~)x. In this case, though, (d/dt)(x?+x~)=2(xf+x~i. This implies that

x?(t)+x~(t)= -1 c2 , c=xi(O)+x~(O). - ct

(7)

Notice that every solution x 1(t), xit) of (7) with xi{O)+ x~(O)#O approaches infinity in finite time. We conclude, therefore, that the equilibrium solution x 1(t)=O, x2(t)=O is unstable. D

Example 1. Consider the system of differential equations

dxl 3 dt = -2x1+x2+3x3+9x2

dx2 5 dt = -6x2-5x3+7x3

dx3 2 2 dt = -x3+xl +x2.

Determine, if possible, whether the equilibrium solution x 1(t)=O, xit)=O, xJCt)=O is stable or unstable. Solution. We rewrite this system in the form x=Ax+g(x) where

[-2

A= ~ 1

-6 0

-~] -1

and g(x)=

9x~

7x~

The function g(x) satisfies the hypotheses of Theorem 2, and the eigenvalues of Aare -2, -6 and -1. Hence, the equilibrium solution x(t)=O is asymptotically stable.

Theorem 2 can also be used to determine the stability of equilibrium solutions of arbitrary autonomous differential equations. Let x0 be an equilibrium value of the differential equation

x=f(x) (8)

389


and set z(t)=x(t)-x0• Then

:i=x=f{x0 +z). (9)

Clearly, z(t)=O is an equilibrium solution of (9) and the stability of x(t)= x0 is equivalent to the stability of z(t):=O.

Next, we show that f(x0 +z) can be written in the form f(x0 +z)=Az+ g(z) where g(z) is small compared to z.

Lemma 1. Let f(x) have two continuous partial derivatives with respect to each of its variables x 1, •.• ,xn. Then, f(x0 + z) can be written in the form

f(x0 + z) = f(x0) + Az + g(z) (10)

where g(z)jmax{lz1!, ... ,1znl} is a continuousfunction ofz which vanishes for z=O.

PROOF # 1. Equation (10) is an immediate consequence of Taylor's Theorem which states that each component Jj(x0 + z) of f(x0 + z) can be written in the form

aij (x0) aij (x0)

Jj(x0 +z)= Jj(x0)+ -a-z 1 + ... + -a-zn+g/z) XI Xn

where ~(z)jmax{iz 1 1, ... , izni} is a continuous function of z which vanishes for z = 0. Hence,

f(x0 + z) = f(x0) + Az + g(z)

where

ajl (xo) ajl (xo)

axl axn

A= 0 ajn (x0) ajn (xo)

axl axn

PRooF #2. If each component of f(x) is a polynomial (possib1y infinite) in x 1, ••• , xn, then each component of f(x0 + z) is a po1ynomia1 in z 1, ••• , zn. Thus,

Jj (x0 +z) = ~0 + a11 z1 + ... + a1nzn + ~(z) (11)

where ~(z) is a po1ynomia1 in z1, ••• ,zn beginning with terms of order two. Setting z=O in (11) gives Jj(x~ = a10. Hence,

f(x0 + z) = f(x0) + Az + g(z),

390


and each component of g(z) is a polynomial in z1, ••• ,zn beginning with terms of order two. D

Theorem 2 and Lemma I provide us with the following algorithm for determining whether an equilibrium solution x(t)=x0 of i=f(x) is stable or unstable:

I. Set z=x-x0•

2. Write f(x0 + z) in the form Az + g(z) where g(z) is a vector-valued polynomial in z1, ••• ,zn beginning with terms of order two or more.

3. Compute the eigenvalues of A. If all the eigenvalues of A have negative real part, then x(t)=x0 is asymptotically stable. If one eigenvalue of A has positive real part, then x( t) = x0 is unstable.

Example 2. Find all equilibrium solutions of the system of differential equations

dx dt =I-xy,

dy 3 -=x-y dt

and determine (if possible) whether they are stable or unstable.

(12)

Solution. The equations I - xy = 0 and x-y 3 = 0 imply that x = I, y = I or x =-I, y = -1. Hence, x(t)= I, y(t)= I, and x(t)= -I, y(t)=- I are the oniy equiiibrium soiutions of (I2).

(i) x(t)= I,y(t)= I: Set u=x-I, v=y-1. Then, du dx dt = dt = I - (I + u )(I +V)= - u- V- uv

~~ = ~ =(I+u)-(I+v)3=u-3v-3v2 -v3•

We rewrite this system in the form

The matrix

(-: -I) -3

has a single eigenvalue A = -2 since

d ( -I-A. et I

Hence, the equilibrium solution x(t)= I, y(t)= I of (I2) is asymptotically stable.

(ii) x(t)= -I,y(t)= -I: Set u=x+ I, v=y+ I. Then, du dx dt = dt = I - ( u- I)( V- I) = u + V - uv

dv dy 3 2 3 - =- =(u-I)-(v-I) = u-3v+3v -v dt dt .

391



~(~)=( ~ _j)(~)+(3~~vv3 ).

The eigenvalues of the matrix

are ;\1 = - 1- V5 , which is negative, and ;\2 = - 1 + V5 , which is positive. Therefore, the equilibrium solution x(t)= -1, y(t)= -1 of (12) is unstable.

Example 3. Find all equilibrium solutions of the system of differential equations

dx . ( ) dt =sm x+y, dy - =ex-1 dt

and determine whether they are stable or unstable.

(13)

Solution. The equilibrium points of (13) are determined by the two equations sin(x+ y)=O and ex -1 =0. The second equation implies that x=O, while the first equation implies that x + y = mr, n an integer. Consequently, x(t)=:O, y(t)= mr, n = 0, ± 1, ± 2, ... , are the equilibrium solutions of (13). Setting u = x, v = y- n'7T, gives

du . ( ) dt = sm u + v + n'7T , ~~ =e"-1.

Now, sin(u + v + n77)=cosn'7Tsin(u + v)=( -ltsin(u + v). Therefore,

~~ = ( -l)nsin(u+ v), ~~ = e" -1.

Next, observe that

(u+v)3

sin(u+v)=u+v- 3, + ... , u2

e" -1 = u + 2! + ....

Hence,

du n [ ( U +V )3 l

dt = (- 1) ( U + V) - 3! + .. . '


fi(u)=((-l)n dt V 1

( -l)n )( ~) + terms of order 2 or 0 higher in u and v.

The eigenvalues of the matrix

((-/)n (-Ol)n)

392


are

When n is even, A1 =(1- V5 )/2 is negative and A2 =(1 + V5 )/2 is positive. Hence, x(t)=O,y(t)=mr is unstable if n is even. When n is odd, both A1 = (- 1- V3 i) /2 and A2 = (- 1 + V3 i) /2 have negative real part. Therefore, the equilibrium solution x( t) = 0, y (t) = mr is asymptotically stable if n is odd.

EXERCISES

Find all equilibrium solutions of each of the following systems of equations and determine, if possible, whether they are stable or unstable.

I. x=x-x3-xy2 j=2y-ys-yx4

4. x=6x-6x2-2xy j=4y-4y2-2xy

2 • .X=x2+y2-J j=x2-y2

5. x=tan(x+y) j=x+x3

3. x=x2+ y 2 -1 j=2xy

6. x=eY-x j=ex+y

Verify that the origin is an equilibrium point of each of the following systems of equations and determine, if possible, whether it is stable or unstable.

7. x=y+3x2 j=x-3y2

10. x=Jn(1 +X+ y 2)

j= -y+x3

12. x=8x-3y+eY-1 j=sinx2-1n(1- x-y)

14. x=x-y+z2 j=y+z-x2 i=z-x+y2

16 • .X=ln(1-z) j=ln(1-x) i=ln(1-y)

8. x=y +cosy -1 j= -sinx+x3

9.x=ex+y_1 j=sin(x+y)

II • .X=cosy-sinx-1 j=x-y-y2

13. x= -x-y-(x2+y2)312 j= x-y +(x2+ y2)3/2

15. x= ex+y+z -1 j=sin(x+y+z) i=x-y-z2

17. x=x-cosy-z+ 1 j=y-cosz-x+ 1 i=z-cosx-y+1

18 (a) Findall equilibrium solutions of the system of differential equations

dx -=gz-hx dt '

dy c dt = a+bx -ky,

dz - =ey-jz. dt .

(fhis system is a model for the contro1 of protein synthesis.) (b) Determine the stability or instability of these solutions if either g, e, or c is

zero.

393


4.4 The phase-plane

In this section we begin our study of the "geometric" theory of differential equations. For simplicity, we will restriet ourselves, for the most part, tc the case n = 2. Our aim is to obtain as complete a description as possible ol all solutions of the system of differential equations

~~ =J(x,y), dy dt =g(x,y). {1)

Tothis end, observe that every solution x=x(t),y=y(t) of (I) defines a curve in the three-dimensional space t,x,y. That is to say, the set of all points (t,x(t),y(t)) describe a curve in the three-dimensional space t,x,y. For example, the solution x = cos t, y = sin t of the system of differential equations

dx dy dt = -y, -=x

dt

describes a helix (see Figure 1) in (t,x,y) space. The geometric theory of differential equations begins with the important

observation that every solution x=x(t),y=y(t), t0 <. t<. 11, of (1) also defines a curve in the x-y plane. To wit, as t runs from t0 to t 1, the set of points (x(t),y(t)) trace out a curve C in the x- y plane. This curve is called the orbit, or trajectory, of the solution x = x(t),y = y(t), and the x-y plane is called the phase-plane of the solutions of (1). Equivalently, we can think of the orbit of x(t), y(t) as the path that the solution traverses in the x-y plane.

y

Figure 1. Graph of the solution x = cos t, y = sin t

Example 1. It is easily verified that x = cos t, y = sin t is a solution of the system of differential equations x =-y, y = x. As t runs from 0 to 277, the set of points (cos t, sin t) trace out the unit circle x2 + y 2 = 1 in the x-y plane. Hence, the unit circle x 2 + y 2 = 1 is the orbit of the solution x=cost, y=sint, 0<. t<.2TT. Ast runs from 0 to oo, the set of points (cost,sint) trace out this circle infinitely often.

394

4.4 The phase-plane

Example 2. It is easily verified that x = e- 1 cost, y = e- 1 sint, - oo < t < oo, is a solution of the system of differential equations dx I dt = - x-y, dy I dt =x-y. Astruns from -oo to oo, the set of points (e- 1 cost,e- 1 sint) trace out a spiral in the x - y plane. Hence, the orbit of the solution x = e- 1 cos t, y = e- 1 sin t is the spiral shown in Figure 2.

y

Figure 2. Orbit of x = e- 1 cost, y = e- 1 sint

Example 3. It is easily verified that x = 3t + 2, y = 5t + 7, - oo < t < oo is a solution of the system of differential equations dx I dt = 3, dy I dt = 5. As t runs from - oo to oo, the set of points (3t + 2, 5t + 7) trace out the straight line through the point (2, 7) with slope t· Hence, the orbit of the solution x=3t+2,y=5t+7 is the straight liney=i(x-2)+7, -oo<x<oo.

Example 4. It is easily verified that x=3t2 +2, y=5t2 +7, O.;;; t<oo is a solution of the system of differential equations

~~ =6[(y-7)1sr12, ~ =w[<x-2)13r12.

All of the points (3t2 +2,5t2 +7) lie on the line through (2,7) with slope t· However, x is always greater than or equal to 2, and y is always greater than or equal to 7. Hence, the orbit of the solution x=3t2+2,y=5t2 +7, 0.;;;; t < oo, is the straight line y = 1<x- 2)+ 7, 2.;;;; x < oo.

Example 5. lt is easily verified that x = 3t + 2, y = 5t2 + 7, - oo < t < oo, is a solution of the system of differential equations

dy 10 dt = T(x- 2)·

395


The orbit of this solution is the set of all points (x,y) = (31 + 2, 512 + 7). Solving for l=~(x-2), we see thaty= ~(x-2f+7. Hence, the orbit of the solution x =31+2, y =512 + 7 is the parabola y = ~(x- 2)2+7, lxl < oo.

One of the advantages of considering the orbit of the solution rather than the solution itself is that is is often possible to obtain the orbit of a solution without prior knowledge of the solution. Let x = x( 1), y = y( I) be a solution of (1). If x'(t) is unequal to zero at t= t1, then we can solve for I= t(x) in a neighborhood of the point x1 = x(t1) (see Exercise 4). Thus, for I near 11• the orbit of the solution x(t), y(l) is the curve y = y(l(x)). Next, observe that

dy dy dt dyjdt g(x,y) dx = dl dx = dx/ dt = f(x,y) ·

Thus, the orbits of the solutions x = x(t), y = y(l) of (I) are the solution curves of the first-order scalar equation

dy g(x,y)

dx = j(x,y) · {2)

Therefore, it is not necessary to find a solution x(l), y(t) of (1) in order to compute its orbit; we need only solve the single first-order scalar differential equation (2).

Remark. From now on, we will use the phrase "the orbits of (I)" to denote the totality of orbits of solutions of (I).

Example 6. The orbits of the system of differential equations

dx _ 2 dy =x2 dl -y' dt {3)

are the solution curves of the scalar equation dy / dx = x2 jy2• This equation is separable, and it is easily seen that every solution is of the formy(x)= (x3 - c)113, c constant. Thus, the orbits of (3) are the set of all curves y = (x3 _ c)I/3.

Example 7. The orbits of the system of differential equations

dx =y(I+x2+y2) dy = -2x(l+x2 +y2) {4) dt ' dl

are the solution curves of the scalar equation

dy 2x(I + x2+ y2) 2x dx =- y(I +x2+y2) =- -y

This equation is separable, and all solutions are of the form iY2 + x2 = c2•

Hence, the orbits of (4) are the families of ellipses iy2+x2=c2.

396

4.4 The phase-plane

Warning. A solution curve of (2) is an orbit of (l) only if dx I dt and dy I dt are not zero simultaneously along the solution. If a solution curve of (2) passes through an equilibrium point of (1), then the entire solution curve is not an orbit. Rather, it is the union of several distinct orbits. For example, consider the system of differential equations

dx = y(l- x2-y2) dy =- x(I- x2-y2). (5) dt ' dt

The solution curves of the scalar equation

dy dyldt X

dx = dx I dt = - y

are the family of concentric circles x 2 + y 2 = c2. Observe, however, that every point on the unit circle x 2 + y 2 = 1 is an equilibrium point of (5). Thus, the orbits of this system are the circles x 2 +y2 =c2, for c"el, and all points on the unit circle x 2 + y 2 = 1. Similarly, the orbits of (3) are the curves y = (x3 -c)113, c"eO; the half-linesy=x, x>O, andy=x, x<O; and the point (0,0).

It is not possible, in general, to explicitly solve Equation (2). Hence, we cannot, in general, find the orbits of (1). Nevertheless, it is still possible to obtain an accurate description of all orbits of (1). This is because the system of differential equations (1) sets up a direction field in the x-y plane. That is to say, the system of differential equations (1) tells us how fast a solution moves along its orbit, and in what direction it is moving. More precisely, Iet x = x(t), y = y(t) be a solution of (1). As t increases, the point (x(t),y(t)) moves along the orbit of this solution. Its velocity in the x-direction is dx I dt; its velocity in the y-direction is dy I dt; and the magnitude of its velocity is [(dx(t)l dti +(dy(t)l dt)2] 112• But dx(t)l dt = f(x(t), y(t)), and dy(t)l dt = g(x(t), y(t)). Hence, at each point (x,y) in the phase plane of (1) we know (i), the tangent to the orbit through (x,y) (the line through (x, y) with direction numbers f(x, y), g(x, y) respectively) and (ii), the speed [J 2(x, y)+ g 2(x, y)] 112 with which the solution is traversing its orbit. As we shall see in Sections 4.8-13, this information can often be used to deduce important properties of the orbits of (1).

The notion of orbit can easily be extended to the case n > 2. Let x = x( t) be a solution of the vector differential equation

ft (X I' · · · 'xn)

x=f(x), f(x)= (6)

on the interval t0 < t < t 1• As t runs from t0 to t 1, the set of points (x 1(t), ... ,xn(t)) trace out a curve C in the n-dimensional space x 1,x2, ••• ,xn. This curve is called the orbit of the solution x=x(t), for t0 ~ t < t 1, and the n-dimensional space x 1, ••• , xn is called the phase-space of the solutions of (6).

397


EXERCISES

In each of Problems l-3, verify that x(t),y(t) is a solution of the given system of equations, and find its orbit.

1. x = l, y = 2(1- x)sin(l- x)2

x(t)=l+t, y(t)=cost2

2. x=e-X, y=e·X-( x(t)=1n(1+t), y(t)=e 1

3. x=1+x2, y=(1+x2)sec2x x(t)=tant, y(t)=tan(tant)

4. Suppose that x'(t1)7to0. Show that we can solve the equation x=x(t) fort= t(x) in a neighborhood of the point x 1 =x(t1). Hint: If x'(t1)7to0, then x(t) is a strictly monotonic function of t in a neighborhood of t = t 1•

Find the orbits of each of the following systems.

s. x=y, y=-x

7. x=y(1 +x+y), y= -x(1 +x+y)

9. x=xye- 3x,

y= -2xy2

11. x=ax-bxy, y=cx-dxy (a,b,c,d positive)

13. x=2xy, y=x2-y2

6. x= y(1 + x 2+ y 2), y= -x(1 +x2+ y 2)

8. x=y+x:Y, y=3x+xy2

10. x=4y, y=x+xy2

12. x = x 2 + cosy, y= -2xy

14. x=y+sinx, y=x-ycosx

4.5 Mathematical theories of war

4.5.1. L. F. Richardson's theory of conflict

In this section we construct a mathematical model which describes the relation between two nations, each determined to defend itself against a possible attack by the other. Each nation considers the possibility of attack quite real, and reasonably enough, bases its apprehensions on the readiness of the other to wage war. Our model is based on the work of Lewis Fry Richardson. It is not an attempt to make scientific Statements about foreign politics or to predict the date at which the next war will break out. This, of course, is clearly impossible. Rather, it is a description of what people would do if they did not stop to think. As Richardson writes: "Why are so many nations reluctantly but steadily increasing their armaments as if they were mechanically compelled to do so? Because, I say, they follow their traditions which are fixtures and their instincts which are mechanical;

398


and because they have not yet made a sufficiently strenuous intellectual and moral effort to control the situation. The process described by the ensuing equations is not to be thought of as inevitable. lt is what would occur if instinct and tradition were allowed to act uncontrolled."

Let x = x ( t) denote the war potential, or armaments, of the first nation, which we will call Jedesland, and Iet y(t) denote the war potential of the second nation, which we will call Andersland. The rate of change of x(t) depends, obviously, on the war readiness y(t) of Andersland, and on the grievances that Jedesland feels towards Andersland. In the most simplistic model we represent these terms by ky and g respectively, where k and g are positive constants. These two terms cause x to increase. On the other hand, the cost of armaments has a restraining effect on dx/ dt. We represent this term by - ax, where a is a positive constant. A similar analysis holds for dy / dt. Consequently, x = x(t), y = y(t) is a solution of the linear system of differential equations

dx dt =ky-ax+g,

dy dt =lx-ßy+h. (I)

Remark. The model (l) is not limited to two nations; it can also represent the relation between two alliances. For example, Andersland and Jedesland can represent the alliances of France with Russia, and Germany with Austria-Hungary during the years immediately prior to World War I.

Throughout history, there has been a constant debate on the cause of war. Over two thousand years ago, Thucydides claimed that armaments cause war. In his account of the Peloponnesian war he writes: "The real though unavowed cause I believe to have been the growth of Athenian power, which terrified the Lacedaemonians and forced them into war." Sir Edward Grey, the British Foreign Secretary during World War I agrees. He writes: "The increase of armaments that is intended in each nation to produce consciousness of strength, and a sense of security, does not produce these effects. On the contrary, it produces a consciousness of the strength of other nations and a sense of fear. The enormous growth of armaments in Europe, the sense of insecurity and fear caused by them-it was these that made war inevitable. This is the real and final account of the origin of the Great War."

On the other hand, L. S. Amery, a member of Britain's parliament during the 1930's vehemently disagrees. When the opinion of Sir Edward Grey was quoted in the House of Commons, Amery replied: "With all due respect to the memory of an eminent statesman, I believe that statement to be entirely mistaken. The armaments were only the symptoms of the conflict of ambitions and ideals, of those nationalist forces which created the War. The War was brought about because Serbia, ltaly and Rumania passionately desired the incorporation in their states of territories which at

399


that time belonged to the Austrian Empire and which the Austrian govemment was not prepared to abandon without a struggle. France was prepared, if the opportunity ever came, to make an effort to recover AlsaceLorraine. lt was in those facts, in those insoluble conflicts of ambitions, and not in the armaments themselves, that the cause of the War lay."

The system of equations (1) takes both conflicting theories into account. Thucydides and Sir Edward Grey would take g and h small compared to k and /, while Mr. Amery would take k and I small compared to g and h.

The system of equations (1) has several important implications. Suppose that g and h are both zero. Then, x(t)=O, y(t)=O is an equilibrium solution of (1). That is, if x, y, g, and h are all made zero simultaneously, then x(t) and y(t) will always remain zero. This ideal condition is permanent peace by disarmament and satisfaction. lt has existed since 1817 on the border between Canada and the United States, and since 1905 on the border between Norway and Sweden.

These equations further imply that mutual disarmament without satisfaction is not permanent. Assurne that x and y vanish simultaneously at some time t = t0• At this time, dx I dt = g and dy I dt = h. Thus, x and y will not remain zero if g and h are positive. Instead, both nations will rearm.

Unilateral disarmament corresponds to setting y = 0 at a certain instant of time. At this time, dy I dt = lx + h. This implies that y will not remain zero if either h or x is positive. Thus, unilateral disarmament is never permanent. This accords with the historical fact that Germany, whose army was reduced by the Treaty of Versailles to 100,000 men, a Ievel far below that of several of her neighbors, insisting on rearming during the years 1933-36.

A race in armaments occurs when the "defense" terms predominate in (1). In this case,

dx dt=ky, dy dt = lx. (2)

Every solution of (2) is of the form

y(t)=Vf [Aevk/ 1 -Be-vk/ 1 ].

Therefore, both x(t) and y(t) approach infinity if A is positive. This infinity can be interpreted as war.

Now, the system of equations (1) is not quite correct, since it does not take into effect the cooperation, or trade, between Andersland and Jedesland. As we see today, mutual cooperation between nations tends to decrease their fears and suspicions. We correct our model by changing the meaning of x(t) and y(t); we let the variables x(t) and y(t) stand for "threats" minus "cooperation." Specifically, we set x = U- U0 and y = V- V0, where U is the defense budget of Jedes1and, V is the defense budget of Andersland, U0 is the amount of goods exported by Jedesland to

400


Andersland and V0 is the amount of goods exported by Andersland to Jedesland. Observe that cooperation evokes reciprocal cooperation, just as armaments provoke more armaments. In addition, nations have a tendency to reduce cooperation on account of the expense which it involves. Thus, the system of equations (I) still describes this more general state of affairs.

The system of equations (I) has a single equilibrium solution

kh+ßg x = Xo = aß- kl '

lg+ah y=yo= aß-kl (3)

if aß- kr=l= 0. We are interested in determining whether this equilibrium solution is stable or unstable. Tothis end, we write (I) in the form w=Aw +f, where

( x(t)) w(t)= y(t) ,

The equilibrium solution is

w=w0 = ( ~:). where Aw0 +f=O. Setting z=w-w0, we obtain that

z =w= Aw + f= A(z +w0) + f= Az+ Aw0 + f=Az.

Clearly, the equilibrium solution w(t)=w0 of w=Aw+f is stable if, and only if, z = 0 is a stable solution of z = Az. To determine the stability of z = 0 we compute

p (;\) = det( - al- A

The roots of p(A) are

_ ; _ ;\ ) = ;\ 2 + ( a + ß );\ + aß- kl.

- ( a + ß) ± [ ( a + ß )2 - 4( aß- kl) J 112

;\= 2

- ( a + ß ) ± [ ( a - ß )2 + 4kl J 112

2 Notice that both roots are real and unequal to zero. Moreover, both roots are negative if aß- kl > 0, and one root is positive if aß- kl < 0. Thus, z( t) =0, and consequently the equilibrium solution x(t)=x0,y(t)=y0 is stable if aß- kl > 0 and unstable if aß- kl < 0.

Let us now tackle the difficult problern of estimating the coefficients a, ß, k, I, g, and h. There is no way, obviously, of measuring g and h. However, it is possible to obtain reasonable estimates for a, ß, k, and /. Observe that the units of these coefficients are reciprocal times. Physicists and engineers would call a - 1 and ß - 1 relaxation times, for if y and g were identically zero, then x(t) = e-a(t-to>x(t0). This implies that x(t0 + a -I)=

401


x(t0)je. Hence, a- 1 is the time required for Jedesland's armaments tobe reduced in the ratio 2.718 if that nation has no grievances and no other nation has any armaments. Richardson estimates a- 1 to be the lifetime of Jedesland's parliament. Thus, a = 0.2 for Great Britain, since the lifetime of Britain's parliament is five years.

To estimate k and l we take a hypothetical case in which g = 0 and y = y 1, so that dxjdt=ky 1-ax. When x=O, 1/k=y1/(dxjdt). Thus, 1/k is the time required for Jedesland to catch up to Andersland provided that (i) Andersland's armaments remain constant, (ii) there are no grievances, and (iii) the cost of armaments doesn't slow Jedesland down. Consider now the German rearmament during 1933-36. Germany started with nearly zero armaments and caught up with her neighbors in about three years. Assuming that the slowing effect of a nearly balanced the Germans' very strong grievances g, we take k = 0.3 (year) -I for Germany. Further, we observe that k is obviously proportional to the amount of industry that a nation has. Thus, k = 0.15 for a nation which has only half the industrial capacity of Germany, and k = 0.9 for a nation which has three times the industrial capacity of Germany.

Let us now check our model against the European arms race of 1909-1914. France was allied with Russia, and Germany was allied with Austria-Hungary. Neither ltaly or Britain was in a definite alliance with either party. Thus, let Jedesland represent the alliance of France with Russia, and let Andersland represent the alliance of Germany with Austria-Hungary. Since these two alliances were roughly equal in size we take k = !, and since each alliance was roughly three times the size of Germany, we take k=/=0.9. We also assume that a=ß=0.2. Then,

dx dy dt = - ax + ky + g, dt = kx- ay + h. ( 4)

Equation (4) has a unique equilibrium point

kh+ag X-o- 2 k2' a -

This equilibrium is unstable since

kg+ah Yo= 2 k2 . a-

aß- kl= a 2 - k2 =0.04-0.81 = -0.77.

This, of course, is in agreement with the historical fact that these two alliances went to war with each other.

Now, the model we have constructed is very crude since it assumes that the grievances g and h are constant in time. This is obviously not true. The grievances g and h are not even continuous functions of time since they jump instantaneously by 1arge amounts. (It's safe to assume, though, that g and h are relatively constant over long periods of time.) In spite of this, the system of equations (4) still provides a very accurate description of the arms race preceding World War I. To demonstrate this, we add the two

402

4.5 Mathematica1 theories of war

equations of (4) together, to obtain that

d dt (X+ y) = ( k- CX )(X+ y) + g + h. (5)

Recall that x = U- U0 and y = V- V0, where U and V are the defense budgets of the two alliances, and U0 and V0 are the amount of goods exported from each alliance to the other. Hence,

The defense budgets for the two alliances are set out in Table I.

Table I. Defense budgets expressed in millions of € sterling

1909 1910 1911 1912 1913

France 48.6 50.9 57.1 63.2 74.7 Russia 66.7 68.5 70.7 81.8 92.0

Germany 63.1 62.0 62.5 68.2 95.4 Austria-Hungary 20.8 23.4 24.6 25.5 26.9

Tota1U+V 199.2 204.8 214.9 238.7 289.0 d(U+ V) 5.6 10.1 23.8 50.3

u + V at same date 202.0 209.8 226.8 263.8

In Figure 1 we plot the annua1 increment of U + V against the average of U + V for the two years used in forming the increment. Notice how close these four points, denoted by o, are to the straight line

d(U+ V)=0.73(U+ V-194). (7)

Thus, foreign politics does indeed have a machine-1ike predictabi1ity. Equations (6) and (7) imply that

g+h=(k-a)(U0 + V0 )-d(U0 + V0 )-194

and k- a = 0.73. This is in excellent agreement with Richardson's estimates of 0.9 for k and 0.2 for a. Finally, observe from (7) that the total defense budgets of the two alliances will increase if U + V is greater than 194 million, and will decrease otherwise. In actual fact, the defense expenditures of the two alliances was 199.2 million in 1909 while the trade between the two alliances amounted to only 171.8 million. Thus began an arms race which led eventually to World War I.

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> 30 + :::)

<J 20

o~L-J---~----~--J---~----~--J----J

190 200 210 220 230 240 250 260 270 u~v

Figure 1. Graph of d( U + V) versus U + V

Rejerence Richardson, L. F ., "Generalized foreign politics," The British Journal of Psychol

ogy, monograph supplement #23, 1939.

ExERCISES

1. Suppose that what moves a government to arm is not the magnitude of other nations' armaments, but the difference between its own and theirs. Then,

'!;; =k(y-x)-ax+g, dy dt =l(x-y)-ßy+h.

Show that every solution of this system of equations is stable if k1/ 1 < (a 1 + k 1)( ß1 + /1) and unstable if k1/ 1 > (a 1 + k1)( ß1 + /1).

2. Consider the case of three nations, each having the same defense coefficient k and the same restraint coefficient a. Then,

dx dt = -ax+ky+kz+g1

dy dt = kx- ay + kz + g2

dz dt =kx+ky-az+g3•

Setting

u=(~)· (-a

A= z k

-a k

we see that u=Au+g. (a) Show thatp(A)=det(A-AI)= -(a+A)3 +3k2(a+A)+2k3•

(b) Show that p(A) = 0 when A = - a- k. Use this information to find the remaining two roots of p(A).

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(c) Show that every solution u(t) is stable if 2k < a, and unstable if 2k > a.

~. Suppose in Problem 2 that the z nation is a pacifist nation, while x and y are pugnacious nations. Then,

dx dt = -ax+ky+kz+gi

: =kx-ay+kz+g2

dz dt =0·x+O·y-az+g3•

(*)

Show that every solution x(t),y(t),z(t) of (*) is stable if k < a, and unstable if k>a.

4.5.2 Lanchester's combat models and the battle of /wo Jima

During the first World War, F. W. Lanchester [4) pointed out the importance of the concentration of troops in modern combat. He constructed mathematical models from which the expected results of an engagement could be obtained. In this section we will derive two of these models, that of a conventional force versus a conventional force, and that of a conventional force versus a guerilla force. We will then solve these models, or equations, and derive "Lanchester's square law," which states that the strength of a combat force is proportional to the square of the number of combatants entering the engagement. Finally, we will fit one of these models, with astonishing accuracy, to the battle of Iwo Jima in World War II.

( a) Construction of the models

Suppose that an "x-force" and a "y-force" are engaged in combat. For simplicity, we define the strengths of these two forces as their number of combatants. (See Howes and Thrall [3] for another definition of combat strength.) Thus Iet x(t) andy(t) denote the number of combatants ofthex and y forces, where t is measured in days from the start of the combat. Clearly, the rate of change of each of these quantities equals its reinforcement rate minus its operationalloss rate minus its combat loss rate.

The operationalloss rate. The operationalloss rate of a combat force is its loss rate due to non-combat mishaps; i.e., desertions, diseases, etc. Lanchester proposed that the operational loss rate of a combat force is proportional to its strength. However, this does not appear to be very realistic. For example, the desertion rate in a combat force depends on a host of psychological and other intangible factors which are difficult even to describe, let alone quantify. We will take the easy way out here and consider only those engagements in which the operationalloss rates are negligible.

The combat /oss rate. Suppose that the x-force is a conventional force which operates in the open, comparatively speaking, and that every member of this force is within "kill range" of the enemy y. We also assume that

405


as soon as the conventional force suffers a loss, fire is concentrated on the remaining combatants. Under these "ideal" conditions, the combat loss rate of a conventional force x equals ay(t), for some positive constant a. This constant is called the combat effectiveness coefficient of the y-force.

The situation is very different if x is a guerilla force, invisible to its opponent y and occupying a region R. The y-force fires into R but cannot know when a kill has been made. It is certainly plausible that the combat loss rate for a guerilla force x should be proportional to x(t), for the !arger x(t), the greater the probability that an opponent's shot will kill. On the other hand, the combat loss rate for x is also proportional to y(t), for the !arger y, the greater the number of x-casualties. Thus, the combat loss rate for a guerilla force x equals cx(t)y(t), where the constant c is called the combat effectiveness coefficient of the Opponent y.

The reinforcement rate. The reinforcement rate of a combat force is the rate at which new combatants enter (or are withdrawn from) the battle. We denote the reinforcement rates of the x- and y-forces by f(t) and g(t) respectively.

Under the assumptions listed above, we can now write down the following two Lanchestrian models for conventional-guerilla combat.

Conventional combat:

Conventional-guerilla combat: ( x = guerilla)

{ dx = -ay+ f(t) dt dy - = -bx+ g(t) dt

{ dx =- cxy+ f(t) dt dy - = -dx+ g(t) dt

(la)

{lb)

The system of equations (la) isalinear system and can be solved explicitly once a, b,f(t), and g(t) are known. On the other hand, the system of equations (lb) is nonlinear, and its solution is much more difficult. (Indeed, it can only be obtained with the aid of a digital computer.)

It is very instructive to consider the special case where the reinforcement rates are zero. This situation occurs when the two forces are "isolated." In this case (la) and (lb) reduce to the simpler systems

dx dt = -ay,

and

dx dt = -cxy,

406

dy -=-bx dt

dy dt = -dx.

(2a)

(2b)


Conventional combat: The square law. The orbits of system (2a) are the solution curves of the equation

dy bx -=-dx ay

dy or ay dx =bx.

Integrating this equation gives

ay 2 - bx2 = ay&- bx&= K. (3)

The curves (3) define a family of hyperbolas in the x-y plane and we have indicated their graphs in Figure 1. The arrowheads on the curves indicate the direction of changing strengths as time passes.

Let us adopt the criterion that one force wins the battle if the other force vanishes first. Then, y wins if K > 0 since the x-force has been annihilated by the time y(t) has decreased to Y K/ a . Similarly, x wins if K<O.

y(t)

Y-K/b x(t)

Figure I. The hyperbolas defined by (3)

Remark 1. Equation (3) is often referred to as "Lanchester's square law," and the system (2a) is often called the square law model, since the strengths of the opposing forces appear quadratically in (3). This terminology is rather anomolous since the system (2a) is actually a linear system.

Remark 2. The y-force always seeks to establish a setting in which K > 0. That is to say, the y-force wants the inequality

ay&> bx&

to hold. This can be accomplished by increasing a; i.e. by using stronger and more accurate weapons, or by increasing the initial force y 0• Notice though that a doubling of a results in a doubling of ay& while a doubling of y 0 results in a jour-fold increase of ay&. This is the essence of Lanchester's square law of conventional combat.

407


Conventional-gueril/a combat. The orbits of system (2b) are the solution curves of the equation

dy = dx =!!:._ dx cxy cy

Multiplying both sides of (4) by cy and integrating gives

cy2 -2dx= cy5-2dx0 = M.

(4)

(5)

The curves (5) define a family of parabolas in the x-y plane, and we have indicated their graphs in Figure 2. The y-force wins if M > 0, since the x

force has been annihilated by the timey(t) has decreased to ~. Similarly, x wins if M < 0.

y (t)

M=O:tie

x(t)

Figure 2. The parabolas defined by (5)

Remark. It is usually impossible to determine, a priori, the numerical value of the combat coefficients a, b, c, and d. Thus, it would appear that Lanchester's combat models have little or no applicability to real-life engagements. However, this is not so. As we shall soon see, it is often possible to determine suitable values of a and b (or c and d) using data from the battle itself. Once these values are established for one engagement, they are known for all other engagements which are fought under similar conditions.

(b) The battle oj !wo Jima

One of the fiercest battles of World War II was fought on the island of lwo Jima, 660 miles south of Tokyo. Our forces coveted lwo Jima as a bornher base close to the Japanese mainland, while the Japanese needed the island as a base for fighter planes attacking our aircraft on their way to bornhing missions over Tokyo and other major Japanese cities. The American inva-

408


sion of lwo Jima began on February 19, 1945, and the fighting was intense throughout the month long combat. Both sides suffered heavy casualties (see Table 1). The Japanese had been ordered to fight to the last man, and this is exactly what they did. The island was declared "secure" by the American forces on the 28th day of the battle, and all active combat ceased on the 36th day. (The last two Japanese holdouts surrendered in 1951!)

Table 1. Casualties at Iwo Jima

Total United States casualties at I wo Jima

Marines Navy units: Ships and air units Medical corpsmen Seabees Doctors and dentists Army units in battle Grand totals

Defense forces (Estimated)

21,000

(Newcomb [6J, page 296)

Killed, missing or Wounded Combat Fatigue died of wounds

5,931

633 195

51 2 9

6,821

17,272

1,158 529 218

12 28

19,217 Japanese casualties at Iwo Jima

Prisoners

Marine 216 Army 867 Total 1,083

2,648

2,648

Total

25,851

1,791 724 269

14 37

28,686

Killed

20,000

The following data is available to us from the battle of lwo Jima. 1. Reinforcement rates. During the conflict Japanese troops were

neither withdrawn nor reinforced. The Americans, on the other band, landed 54,000 troops on the first day of the battle, none on the second, 6,000 on the third, none on the fourth and fifth, 13,000 on the sixth day, and none thereafter. There were no American troops on lwo Jima prior to the start of the engagement.

2. Combat losses. Captain Clifford Morehouse of the United States Marine Corps (see Morehouse [5]) kept a daily count of all American combat losses. Unfortunately, no such records are available for the Japanese forces. Most probably, the casualty lists kept by General Kuribayashi (commander of the Japanese forces on Iwo Jima) were destroyed in the battle itself, while whatever records were kept in Tokyo were consumed in the fire bombings of the remaining five months of the war. However, we can infer from Table I that approximately 21,500 Japanese forces were on lwo Jima at the start of the battle. (Actually, Newcomb arrived at the fig-

409


ure of 21,000 for the Japanese forces, but this is a 1ittle 1ow since he apparently did not include some of the living and dead found in the caves in the final days.)

3. Operationallosses. The operationa11osses on both sides were negligible.

Now, let x(t) and y(t) denote respective1y, the active American and Japanese forces on Iwo Jima t days after the battle began. The data above suggests the following Lanchestrian mode1 for the battle of Iwo Jima:

dx = -ay+j(t) dt dy -=-bx dt

(6)

where a and b are the combat effectiveness coefficients of the Japanese and American forces, respectively, and

54,000 O"t<1 0 1"t<2

f(t)= 6,000 2"t<3 0 3"t<5

13,000 5"t<6 0 1~6

Using the method of variation of parameters deve1oped in Section 3.12 or the method of elimination in Section 2.14, it is easily seen that the solution of (6) which satisfies x(O)=O, y(O)=y0 =21,500 is given by

x(t) =-Vf y0 coshv'lib t + {coshv'lib (t -s)f(s)ds (7a)

and

y(t)=y0 coshv'lib t-Vf {sinhv'lib (t-s)f(s)ds (7b)

where

The problern before us now is this: Do there exist constants a and b so that (7a) yields a good fit to the data compiled by Morehouse? This is an extremely important question. An affirmative answer would indicate that Lanchestrian models do indeed describe real life battles, while a negative answer would shed a dubious light on much of Lanchester's work.

As we mentioned previously, it is extremely difficult to compute the combat effectiveness coefficients a and b of two opposing forces. However, it is often possible to determine suitable values of a and b once the data for the battle is known, and such is the case for the battle of lwo Jima.

410


The calculation of a and b. Integrating the second equation of (6) between 0 and s gives

so that

b= Yo-y(s).

fosx(t)dt

In particular, setting s = 36 gives

b= Yo-y(36) = 21,500 (36 (36

Jo x(t)dt Jo x(t)dt

(8)

(9)

Now the integral on the right-hand side of (9) can be approximated by the Riemann sum

36 36 l x(t)dt:;;t. ~ x(i) 0 i= I

and for x(i) we enter the number of effective American troops on the ith day of the battle. Using the data available from Morehouse, we compute for b the value

21,500 b= 2,037,000 =0.0106. (10)

Remark. We would prefer to set s = 28 in (8) since that was the day the island was declared secure, and the fighting was only sporadic after this day. However, we don't know y(28). Thus, we are forced here to take s=36.

Next, we integrate the first equation of (6) between t=O and t=28 and obtain that

(28 (28 x(28)= -a Jo y(t)dt+ Jo f(t)dt

(28 =- a Jo y(t)dt+73,000.

There were 52,735 effective American troops on the 28th day of the battle. Thus

a = 73,000- 52,735 = 20,265 (28 (28

Jo y(t)dt Jo y(t)dt

(11)

411


Finally, we approximate the integral on the right-hand side of (11) by the Riemann sum

28 28 l y(t)dt~ ~ y(J) 0 j= I

and we approximate y(j) by

y(i)=y0 -b fojx(t)dt

j

~21,500- b ~ x(i). i=i

Again, we replace x(i) by the number of effective American troops on the ith day of the battle. The result of this calculation is (see Engel [2])

20,265 a = 372,500 = 0.0544. (12)

Figure 3 below compares the actual American troop strength with the values predicted by Equation (7a) (with a=0.0544 and b=0.0106). The fit is remarkably good. Thus, it appears that a Lanchestrian model does indeed describe reallife engagements.

(/)

LLI (.J er: 0 LL

z Cl (.J

er: LLI :Ii Cl

74,000

70,000

66,000

62,000

58,000

54,000

50,000

--- ACTUAL -PREDICTED

TROOPS

4 8 12 16 20 24 28 32 36

DAYS

Figure 3. Comparison of actual troop strength with predicted troop strength

Remark. The figures we have used for American reinforcements include a/1 the personnel put ashore, both combat troops and support troops. Thus the numbers a and b that we have computed should be interpreted as the average effectiveness per man ashore.

412


References I. Coleman, C. S., Combat Models, MAA Workshop on Modules in Applied

Math, Cornell University, Aug. 1976. 2. Engel, J. H., A verification of Lanchester's law, Operations Research, 2, (1954),

163-171.

3. Howes, D. R., and Thrall, R. M., A theory of ideallinear weights for heterogeneous combat forces, Naval Research Logislies Quarterly, vol. 20, 1973, pp. 645-659.

4. Lanchester, F. W., Aircraft in Warfare, the Dawn of the Fourth Arm. Tiptree, Constable and Co., Ltd., 1916.

5. Morehouse, C. P., The /wo Jima Operation, USMCR, Historical Division, Hdqr. USMC, 1946.

6. Newcomb, R. F., /wo Jima. New York: Holt, Rinehart, and Winston, 1965.

EXERCISES

1. Derive Equations (7a) and (7b).

2. The system of equations

.X= -ay (13)

y= -by-cxy

is a Lanchestrian model for conventional-guerilla combat, in which the operationalloss rate of the guerilla force y is proportional to y(t). (a) Find the orbits of (13). (b) Who wins the battle?

3. The system of equations .X= -ay

y= -bx-cxy (14)

is a Lanchestrian model for conventional-guerilla combat, in which the operationalloss rate of the guerilla force y is proportional to the strength of the conventional force x. Find the orbits of (14).

4. The system of equations .X= -cxy

y= -dxy (15)

is a Lanchestrian model for guerilla-guerilla combat in which the operational loss rates are negligible. (a) Find the orbits of (15). (b) Who wins the battle?

5. The system of equations .X= -ay-cxy

y= -bx-dxy (16)

is a Lanchestrian model for guerilla-guerilla combat in which the operational loss rate of each force is proportional to the strength of its opponent. Find the orbits of (16).

413


6. The system of equations

x= -ax-c.zy

y= -by-d.zy (17)

is a Lanchestrian model for guerilla-guerilla combat in which the operational loss rate of each force is proportional to its strength. (a) Find the orbits of (17). (b) Show that the x andy axes are both orbits of (17). (c) Using the fact (tobe proved in Section 4.6) that two orbits of (17) cannot intersect, show that there is no clear-cut winner in this battle. Hint: Show that x(t) and y(t) can never become zero in finite time. (Using lemmas l and 2 of Section 4.8, it is easy to show that both x(t) and y(t) approach zero as t-+oo.)

4.6 Qualitative properties of orbits

In this section we will derive two very important properties of the solutions and orbits of the system of differential equations

_ [f1 (~1, ... ,xn)

f(x)- . .

fn(x1, ... ,xn)

x=f(x), (1)

The first property deals with the existence and uniqueness of orbits, and the second property deals with the existence of periodic solutions of (1). We begin with the following existence-uniqueness theorem for the solutions of (1).

Theorem 3. Let each of the functions j 1(x 1, ... ,xn), ... Jn(x 1, ... ,xn) have continuous partial derivatives with respect to x 1, ••• , xn. Then, the initial-value problern x = f(x), x(t0) = x0 has one, and only one solution x = x( t), for every x0 in Rn.

We prove Theorem 3 in exactly the samemanneras we proved the existence-uniqueness theorem for the scalar differential equation .X= f(t,x). Indeed, the proof given in Section 1.10 carries over here word for word. We need only interpret the quantity !x( t)- y( t)l, where x( t) and y( t) are vector-valued functions, as the length of the vector x( t)- y( t). That is to say, if we interpret !x(t)-y(t)i as

lx(t) -y(t)l =max{lx1 (t)-Y 1 (t)i, .. . , lxn(t)-Yn(t)i },

then the proof of Theorem 2, Section 1.10 is valid even for vector-valued functions f(t,x) (see Exercises 13-14).

Next, we require the following simple but extremely usefullemma.

Lemma 1. ljx=q,(t) is a solution of(l), then x=q,(t+c) is again a solution of (1).

414


The meaning of Lemma 1 is the following. Let x=f[>(t) be a solution of (1) and let us replace every t in the formula for f[>(t) by t + c. In this manner we obtain a new function i(t)=f[>(t+ c). Lemma 1 states that x(t) is again a solution of (1). For example, x 1 = tant, x2 =sec2 t is a solution of the system of differential equations dx1/ dt = x2, dx2/ dt = 2x1x2• Hence, x1 =tan(t+c), x2 =sec2(t+c) is again a solution, for any constant c.

PR.ooF OF LEMMA 1. If x=f[>(t) is a solution of (1), then df[>(t)/ dt=f(f[>(t)); that is, the two functions df[>(t)/ dt and h(t)=f(f[>(t)) agree at every single time. Fixa timet and a constant c. Since dq,j dt and h agree at every time, they must agree at time t + c. Hence,

dq, dt(t + c) = h(t + c) =f(q,(t + c)).

But, dq,jdt evaluated at t+c equals the derivative of i(t)=f/>(t+c), evaluated at t. Therefore,

d dt q,(t + c) =f(q,(t + c)). 0

Remark 1. Lemma I can be verified explicitly for the linear equation x = Ax. Every solution x( t) of this equation is of the form x( t) = eA1v, for some constant vector v. Hence,

x(t + c) = eA(t+c)v= eAteAcv

since (At)Ac=Ac(At) for all values oft and c. Therefore, x(t+c) is again a solution of x = Ax since it is of the form eA1 times the constant vector eAcv.

Remark 2. Lemma I is not true if the function f in (I) depends explicitly on t. To see this, suppose that x=f[>(t) is a solution of the nonautonomaus differential equation x = f(t, x). Then, 1>U + c) = f(t + c, f[>(t + c)). Consequently, the function x = f/>( t + c) satisfies the differential equation

x=f(t+ c,x),

and this equation is different from (1) if f depends explicitly on t.

We are now in a position to derive the following extremely important properties of the solutions and orbits of (1).

Property 1. (Existence and uniqueness of orbits.) Let each of the functions / 1(x1,. •• ,xn), ... Jn(x 1, ••• ,xn) have continuous partial derivatives with respect to x1, ••• ,xn. Then, there exists one, and only one, orbit through every point x0 in Rn. In particular, if the orbits of two solutions x=f[>(t) and x=t{;(t) of (I) have one point in common, then they must be identical.

415


Property 2. Let x = cf>( t) be a solution of (I). If cp(t0 + T) = cp(t0) for some t0

and T > 0, then cf>( t + T) is identically equal to cf>( t). In other words, if a solution x( t) of (1) returns to its starting value after a time T > 0, then it must be periodic, with period T (i.e. it must repeat itself over every time interval of length T.)

PROOF OF PROPERTY 1. Let x0 be any point in the n-dimensional phase space x 1, ... ,xn, and Iet x=c[>(t) be the solution of the initial-value problern x = f(x), x(O) = x0. The orbit of this solution obviously passes through x0•

Hence, there exists at least one orbit through every point x0• Now, suppose that the orbit of another solution x = l[i(t) also passes through x0. This means that there exists t0(*0) suchthat l[i(t0)=x0• By Lemma l,

x=l[i(t+t0 )

is also a solution of (I). Observe that l[i(t + t0) and cp(t) have the same value at t = 0. Hence, by Theorem 3, l[i(t + t0) equals cp(t) for all time t. This implies that the orbits of cp(t) and l[i(t) are identical. To wit, if ~ is a point on the orbit of cf>(t); that is, ~=cp(t 1 ) for some t 1, then ~ is also on the orbit of I[!( t), since ~= cf>( t 1) =I[!( t 1 + t0). Conversely, if ~ is a point on the orbit of l[i(t); that is, there exists t2 suchthat l[i(t2)=~, then ~ is also on the orbit of cf>( t) since ~= l[i(t2) = cf>( t2 - t0). 0

PROOF OF PROPERTY 2. Let x=c[>(t) be a solution of (1) and suppose that cf>(t0 + T) = cp(t0) for some numbers t0 and T. Then, the function I[!( t) = cf>( t + T) is also a solution of (1) which agrees with cf>( t) at time t = t0• By Theorem 3, therefore, I[!( t) = cf>( t + T) is identically equal to cf>( t). 0

Property 2 is extremely useful in applications, especially when n = 2. Let x = x(t), y = y(t) be a periodic solution of the system of differential equations

~~ =f(x,y), dy dt =g(x,y). (2)

If x(t + T) = x(t) and y(t + T) = y(t), then the orbit of this solution is a closed curve C in the x-y plane. In every time interval t0 < t < t0 + T, the solution moves once around C. Conversely, suppose that the orbit of a solution x = x(t), y = y(t) of (2) is a closed curve containing no equilibrium points of (2). Then, the solution x = x(t), y = y(t) is periodic. To prove this, recall that a solution x = x(t), y = y(t) of (2) moves along its orbit with velocity [P(x,y)+ g2(x,y)] 112• If its orbit Cis a closed curve containing no equilibrium points of (2), then the function [J2(x,y) + g2(x,y)] 112 has a positive minimum for (x,y) on C. Hence, the orbit of x = x(t), y = y(t) mustreturn to its starting point x 0 =x(t0),y0 =y(t0) in some finite timeT. Butthis implies that x(t+ T)=x(t) andy(t+ T)=y(t) for all t.

Example 1. Prove that every solution z(t) of the second-order differential equation ( d 2z j dt 2) + z + z3 = 0 is periodic.

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PROOF. We convert this second-order equation into a system of two firstorder equations by setting x=z, y=dzldt. Then,

~~ =y, ~ = -x-x3• (3)

The orbits of (3) are the solution curves

y2 x2 x4 2

2+2+4=c (4)

of the scalar equation dyldx= -(x+x3)IY· Equation (4) defines a closed curve in the x-y plane (see Exercise 7). Moreover, the only equilibrium point of (3) is x =O,y =0. Consequently, every solution x = z(t),y = z'(t) of (3) is a periodic function of time. Notice, however, that we cannot compute the period of any particular solution. 0

Example 2. Prove that every solution of the system of differential equa-tions

(5)

is periodic. Solution. The orbits of (5) are the solution curves x 2 + y 2 = c2 of the firstorder scalar equation dy I dx = - x I y. Moreover, x = 0, y = 0 is the only equilibrium point of (5). Consequently, every solution x = x(t), y = y(t) of (5) is a periodic function of time.

EXERCISES

1. Show that all solutions x(t),y(t) of

dx =x2 +ysinx dt '

dy - = -1 +xy+cosy dt

which start in the first quadrant (x > 0, y > 0) must remain there for all time (both backwards and forwards).

2. Show that all solutions x(t),y(t) of

'! =y(ex-1), dy -=x+eY dt

which start in the right half plane (x > 0) must remain there for all time.

3. Show that all solutions x(t), y(t) of dy -=xy+tany dt

which start in the upper half plane (y > 0) must remain there for all time.

4. Show that all solutions x(t), y(t) of

dx 2 -= -1-y+x dt '

dy dt =x+xy

which start inside the unit circle x 2 + y 2 = I must remain there for all time. Hint: Compute d(x 2 +y2)/dt.

417


5. Let x(t), y(t) be a solution of

~~ =y+x2, : =x+y2

with x(t0h~' y(t0). Show that x(t) can never equal y(t).

6. Can a figure 8 ever be an orbit of

dx dt =J(x,y),

dy dt =g(x,y)

where f and g have continuous partial derivatives with respect to x and y?

7. Show that the curvey2+x2+x4/2=2c2 is closed. Hint: Show that there exist two points y = 0, x = ± a which lie on this curve.

Prove that all solutions of the following second-order equations are periodic.

8. d2z +z3=0 dt2

10. d2z + ez2= I dt2

9. d2z +z+zs=o dt2

1 d 2z +-z-=O 1 . 2 dt2 l+z

12. Show that all solutions z(t) of

d2z +z-2z3=0 dt2

are periodic if i 2(0) + z2(0)- z4(0) < ! , and unbounded if

i 2{0)+z2(0)- z4(0) >!·

13. (a) Let

L=nx . . max la.fi/axA, for lx-x0l<:b. IJ-= l, ... ,n

Show that lf(x)-f(y)l < Llx-yl if lx-x01 <band ly-x01 < b. (b) Let M=maxlf(x)l for lx-x01 < b. Show that the Picard iterates

converge to a solution x(t) of the initial-value problern i=f(x), x(t0)=x0 on the interval lt- t0l < b / M. Hint: The proof of Theorem 2, Section 1.10 carries over here word for word.

14. Cornpute the Picard iterates xi(t) of the initial-value problern i=Ax, x(O)-x0,

and verify that they approach eA1x0 asj approaches infinity.

4.7 Phaseportraits of linear systems

In this section we present a complete picture of all orbits of the linear differential equation

i=Ax, x=(~~). A=(~ ~)· (I)

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This picture is called a phase portrait, and it depends almost completely on the eigenvalues of the matrix A. lt also changes drastically as the eigenvalues of A change sign or become imaginary.

When analyzing Equation (1), it is often helpful to visualize a vector

x=(~~) in R2 as a direction, or directed line segment, in the plane. Let

be a vector in R2 and draw the directed line segment x from the point (0, 0) to the point (x1,x2), as in Figure la. This directed line segment is parallel to the line through (0,0) with direction numbers x 1,x2 respectively. If we visualize the vector x as being this directed line segment x, then we see that the vectors x and cx are parallel if c is positive, and antiparallel if c is negative. We can also give a nice geometric interpretation of vector addition. Let x and y be two vectors in R2• Draw the directed line segment x, and place the vector y at the tip of x. The vector x + y is then the composi-

(-2,-1)

(a) (b)

Figure I

Figure 2

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tion of these two directed line segments (see Figure 2). This construction is known as the parallelogram law of vector addition.

We are now in a position to derive the phase portraits of (1). Let A1 and A2 denote the two eigenvalues of A. W e distinguish the following cases.

1. A2 < A1 < 0. Let v1 and v2 be eigenvectors of A with eigenvalues A1 and A2 respectively. In the x 1 - x2 plane we draw the four half-lines /1, /), / 2, and 12, as shown in Figure 3. The rays /1 and /2 areparallel to v1 and v2, while the rays /) and 12 areparallel to -v1 and -v2• Observe first that x(t)= ce>--'1v1 is a solution of (1) for any constant c. This solution is always proportional to v1, and the constant of proportionality, ce>--' 1, runs from ± oo to 0, depending as to whether c is positive or negative. Hence, the orbit of this solution is the half-line /1 for c > 0, and the half-line /) for c < 0. Similarly, the orbit of the solution x(t)=ce>--21v2 is the half-line /2 for c>O, and the half-line 12 for c < 0. The arrows on these four lines in Figure 3 indicate in what direction x(t) moves along its orbit.

Next, recall that every solution x(t) of (1) can be written in the form

x(t)= c1e>..•1v1 + c2e>.. 21v2 (2)

for some choice of constants c1 and c2• Obviously, every solution x(t) of (I) approaches C8) as t approaches infinity. Hence, every orbit of (1) approaches the origin x 1 = x2 = 0 as t approaches infinity. We can make an even stronger statement by observing that e>--21v2 is very small compared to e>--'1v1 when t is very large. Therefore, x(t), for c1 *0, comes closer and closer to c1e>..•1v1 as t approaches infinity. This implies that the tangent to the orbit of x(t) approaches /1 if c1 is positive, and /) if c1 is negative. Thus,

Figure 3. Phase portrait of a stable node

420

4.7 Phase portraits of linear systems

the phase portrait of (1) has the form described in Figure 3. The distinguishing feature of this phase portrait is that every orbit, with the exception of a single 1ine, approaches the origin in a fixed direction (if we consider the directions v1 and - v1 equiva1ent). In this case we say that the equilibrium solution x=O of (l) is a stab1e node.

Remark. The orbit of every solution x(t) of (l) approaches the origin x 1 = x2 = 0 as t approaches infinity. However, this point does not belang to the orbit of any nontrivial solution x( t).

1'. 0<.\1 <.\2• The phase portrait of (1) in this case is exactly the same as Figure 3, except that the direction of the arrows is reversed. Hence, the equi1ibrium solution x(t)=O of (I) is an unstable node if both eigenvalues of A are positive.

2 . .\1 =.\2 <0. In this case, the phase portrait of (1) depends on whether Ahasone or two linearly independent eigenvectors. (a) Suppose that A has two linearly independent eigenvectors v1 and v2 with eigenvalue .\ < 0. In this case, every solution x(t) of (1) can be written in the form

(2)

for some choice of constants c1 and c2. Now, the vector eM(c1v1+c2v2) is parallel to c1v 1 + c2v2 for all t. Hence, the orbit of every solution x(t) of (l) is a half-line. Moreover, the set of vectors { c1v1 + c2v2}, for all choices of c1 and c2, cover every direction in the x 1 - x2 plane, since v1 and v2 are linearly independent. Hence, the phase portrait of (I) has the form described in Figure 4a. (b) Suppose that A has only one linearly independent eigenvector v, with eigenvalue .\. In this case, x1(t) = eMv is one solution of (I). To find a second solution of (I) which is independent of x1, we observe that (A-.\1)2u=O for every vector u. Hence,

(3)

is a solution of (l) for any choice of u. Equation (3) can be simplified by observing that (A- .\l)u must be a multiple k of v. This follows immediately from the equation (A- .\I)[(A- .\I)u] = 0, and the fact that A has only one linearly independent eigenvector v. Choosing u independent of v, we see that every solution x(t) of (1) can be written in the form

x(t) = c1eMv+ c2eM (u+ ktv)= e>..t (c1v+ c2u+ c2ktv), (4)

for some choice of constants c1 and c2• Obviousiy, every solution x(t) of (1) approaches (8) as t approaches infinity. In addition, observe that c1 v + c2u is very small compared to c2ktv if c2 is unequal to zero and t is very large. Hence, the tangent to the orbit of x(t) approaches ± v (depending on the sign of c2) as t approaches infinity, and the phase portrait of (1) has the form described in Figure 4b.

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(a) (b)

Figure 4

2'. A1 =A2 >0. The phase portraits of (1) in the cases (2a)' and (2b)' are exactly the same as Figures 4a and 4b, except that the direction of the arrows is reversed.

3. A1 <0<A2• Let v1 and v2 be eigenvectors of A with eigenvalues A1 and A2 respectively. In the x 1 - x2 plane we draw the four half-lines 11, 1{, 12, and I~; the half-lines 11 and 12 are parallel to v1 and v2, while the half-lines I{ and /~ areparallel to -v1 and -v2• Observe first that every solution x(t) of (f) is of the form

(5)

for some choice of constants c1 and c2• The orbit of the solution x(t)= c 1 e~' 1v 1 is /1 for c1 > 0 and /{ for c1 < 0, while the orbit of the solution x(t)=c2e~21V2 is /2 for c2 >0 and /~ for c2 <0. Note, too, the direction of the arrows on /1, /{, 12, and /~; the solution x(t)=c 1 e~' 1V 1 approaches (8) as t approaches infinity, where.as the solution x(t) = c2e~21v2 becomes unbounded (for c2 =FO) as t approaches infinity. Next, observe that e~' 1v 1 is very small compared to e~21V2 when t is very large. Hence, every solution x(t) of (1) with c2 =FO becomes unbounded as t approaches infinity, and its orbit approaches either /2 or /~. Finally, observe that e~21v2 is very small compared to e~' 1v 1 when t is very large negative. Hence, the orbit of any solution x(t) of (1), with c1 =FO, approaches either /1 or /{ as t approaches minus infinity. Consequently, the phase portrait of (I) has the form described in Figure 5. This phase portrait resembles a "saddle" near x 1 = x2 =

422


Figure 5. Phaseportrait of a saddle point

0. Forthis reason, we say that the equilibrium solution x(t)=O of (l) is a saddle point if the eigenvalues of A have opposite sign.

4. A1 = a + iß, A2 = a- iß, ß =I= 0. Our first step in deriving the phase portrait of (1) is to find the general solution of (1). Let z=u+iv be an eigenvector of A with eigenvalue a + iß. Then,

x{t) = e<a+iß)t(u + iv) = ea1 (cosßt + i sinßt)(u + iv)

= ea1 [ ucosßt -vsinßt] + ie"1 [ usinßt+vcosßt]

is a complex-valued solution of (1). Therefore,

x1{t) = ea1[ ucosßt -vsinßt]

and

x2(t) = e"1[ usinßt+vcosßt]

are two real-valued linearly independent solutions of (1), and every solution x(t) of (l) is of the form x(t)=c1x1(t)+c2x2(t). This expression can be written in the form (see Exercise 15)

x(t)=eat(Ricos(ßt-81)) {6) R2 cos{ ßt - 82 )

for some choice of constants R1 > 0, R2 > 0, 81, and 82• We distinguish the following cases.

(a) a=O: Observe that both

x1 (t)= R1 cos(ßt- 81 ) and xit) = R2 cos( ßt- 82 )

are periodic functions of time with period 27T / ß. The function x 1(t) varies between - R1 and + R1, while x2(t) varies between - R2 and + R2• Conse-

423


quently, the orbit of any solution x(t) of (1) is a closed curve surrounding the origin x1 =x2 =0, and the phase portrait of (1) has the form described in Figure 6a. Forthis reason, we say that the equilibrium solution x(t)=O of (1) is a center when the eigenvalues of Aare pure imaginary.

The direction of the arrows in Figure 6a must be determined from the differential equation (1). The simplest way of doing this is to check the sign of .X2 when x2 = 0. If .X2 is greater than zero for x2 = 0 and x 1 > 0, then all so1utions x(t) of (1) move in the counterc1ockwise direction; if .X2 is less than zero for x2 =0 and x1 >0, then all solutions x(t) of (1) move in the clockwise direction.

{a) {b)

(c)

Figure 6. (a) a=O; (b) a<O; (c) a>O

424


(b) a<O: In this case, the effect of the factor ea1 in Equation (6) is to change the simple closed curves of Figure 6a into the spirals of Figure 6b. This is because the point x(27T / ß) = e2wa/ ßx(O) is closer to the origin than x(O). Again, the direction of the arrows in Figure 6b must be determined directly from the differential equation (1). In this case, we say that the equilibrium solution x(t)=O of (1) is a stable focus.

(c) a > 0: In this case, all orbits of (1) spiral away from the origin as t approaches infinity (see Figure 6c), and the equilibrium solution x(t)=O of (1) is called an unstable focus.

Finally, we mention that the phase portraits of nonlinear systems, in the neighborhood of an equilibrium point, are often very similar to the phase portraits of linear systems. More precisely, let x = x0 be an equilibrium solution of the nonlinear equation x = f(x), and set u = x- x0• Then, (see Section 4.3) we can write the differential equation x=f(x) in the form

ü=Au+g(u) (7)

where A is a constant matrix and g(u) is very small compared to u. We state without proof the following theorem.

Theorem 4. Suppose that u=O is either a node, saddle, or focus point of the differential equation ü =Au. Then, the phase portrait of the differential equation x = f(x), in a neighborhood of x = x0, has one of the forms described in Figures 3, 5, and 6 (band c), depending as to whether u=O is a node, saddle, or focus.

Example 1. Draw the phase portrait of the linear equation

. Ax ( -2 X= = 4 -1) -7 X. (8)

Solution. It is easily verified that

v1={ D and v2={!) are eigenvectors of A with eigenvalues -3 and -6, respectively. Therefore, x = 0 is a stable node of (8), and the phase portrait of (8) has the form described in Figure 7. The half-line /1 makes an angle of 45° with the x1

axis, while the half-line 12 makes an angle of 0 degrees with the x 1-axis, where tan 0 = 4.


x=Ax=( I -3

Solution. It is easily verified that

-3) I X.

v1={ D and v2={- D

(9)

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Figure 7. Phase portrait of (8)


are eigenvectors of A with eigenvalues -2 and 4, respectively. Therefore, x = 0 is a saddle point of (9), and its phase portrait has the form described in Figure 8. The half-line /1 makes an angle of 45° with the x 1-axis, and the half-line /2 is at right angles to /1•


x = Ax = ( - 1 1 )x. -1 -1

(10)

426

4.7 Phase portraits of linear systems

Solution. The eigenvalues of A are - 1 ± i. Hence, x = 0 is a stable focus of (10) and every nontrivial orbit of (10) spirals into the origin as t approaches infinity. To determine the direction of rotation of the spiral, we observe that i 2 = - x 1 when x 2 = 0. Thus, i 2 is negative for x 1 > 0 and x2 = 0. Consequently, all nontrivial orbits of (10) spiral into the origin in the clockwise direction, as shown in Figure 9.

Xz


EXERCISES

Draw the phase portraits of each of the following systems of differential equations.

1 . (-5 1) . x= 1 -5 x 2. x=(~ -1) -6 X 3. x=( -~ -1) 5 X

4. x=( -1 -1) -6 X s. x=( -~ -4) 4 X 6. x=(~ -1) -3 X

7. x=(-~ 2)x -1 s. x=O -1) -3 X 9. x=( _; 1 )x

-2

10. Show that every orbit of

x=( o -9 ci )x

is an ellipse.

11. The equation of motion of a spring-mass system with damping (see Section 2.6) is mi + ci + kz = 0, where m, c, and k are positive numbers. Convert this equation to a system of first-order equations for x=z, y=i, and draw the phase portrait of this system. Distinguish the overdamped, critically damped, and underdamped cases.

12. Suppose that a 2 X 2 matrix A has 2 1inearly independent eigenvectors with eigenvalue A.. Show that A ='Al

427


13. This problern illustrates Theorem 4. Consider the system

~~ =y, : =x+2x3• (*)

(a) Show that the equilibrium solution x=O,y=O of the linearized system x= y, j = x is a saddle, and draw the phase portrait of the linearized system.

(b) Find the orbits of (*), and then draw its phase portrait. (c) Show that there are exactly two orbits of (*) (one for x >0 and one for x <

0) on which x-+0, y-+0 as t-+oo. Similarly, there are exactly two orbits of (*) on which x-+0, y-+0 as t-+- oo. Thus, observe that the phase portraits of (*) and the linearized system Iook the same near the origin.

14. Verify Equation (6). Hint: The expression acos"'t+bsin"'t can always be written in the form Rcos("'t-8) for suitable choices of Rand 8.

4.8 Long time behavior of solutions; the Poincare-Bendixson Theorem

We consider now the problern of determining the long time behavior of all solutions of the differential equation

x=f(x), (I)

This problern has been solved completely in the special case that f(x) = Ax. As we have seen in Sections 4.2 and 4.7, all solutions x(t) of x=Ax must exhibit one of the following four types of behavior: (i) x(t) is constant in time; (ii) x(t) is a periodic function of time; (iii) x(t) is unbounded as t approaches infinity; and (iv) x(t) approaches an equilibrium point as t approaches infinity.

A partial solution to this problem, in the case of nonlinear f(x), was given in Section 4.3. In that section we provided sufficient conditions that every solution x(t) of (1), whose initial value x(O) is sufficiently close to an equilibrium point ~. must ultimately approach ~ as t approaches infinity. In many applications it is often possible to go much further and prove that every physically (biologically) realistic solution approaches a single equilibrium point as time evolves. In this context, the following two Iemmas play an extremely important role.

Lemma 1. Let g(t) be a monotonic increasing (decreasing) function of time fort> t0, with g(t) < c( > c) for some constant c. Then, g(t) has a Iimit as t approaches infinity.

PRooF. Suppose that g(t) is monotonic increasing for t > t0, and g(t) is bounded from above. Let I be the least upper bound of g; that is, I is the smallest number which is not exceeded by the values of g(t), fort;> t0• This

428

4.8 Long time behavior of solutions: the Poincarb-Bendixson Theorem

number must be the Iimit of g(t) as 1 approaches infinity. To prove this, let E > 0 be given, and observe that there exists a time 1. ~ 10 such that /- g( 1.)

< E. (If no such time 1. exists, then I is not the least upper bound of g.) Since g( t) is monotonic, we see that /- g( I)< E for t ~ t •. This shows that /=limH00 g(l). D

Lemma 2. Suppose lhal a solulion x(t) of (1) approaches a veclor ~ as I ap-proaches infinity. Then, ~ is an equilibrium poinl of (1).

PROOF. Suppose that x(t) approaches ~ as 1 approaches infinity. Then, ~(I) approaches ~. where ~ is the jth component of ~· This implies that lx1(t1)- x1 (t2)1 approaches zero as both 11 and 12 approach infinity, since

lxi 11)- xi 12)1 =I( x1( 11)- ~J) + (~1 - x1( t2))1

~ lxAtl) -~1 1 + lxA12)-~I· In particular, let t 1 = t and t2 = t 1 + h, for some fixed positive number h. Then, lx/t + h)- x/t)l approaches zero as t approaches infinity. But

dx/T) x1(t + h)- x1(t) = h--;[1 = hjj (x 1 ( T), ... ,xn(T)),

where r is some number between t and t + h. Finally, observe that jj(x 1(T), ... ,xn(T)) must approach jj(~p ... '~n) as t approaches infinity. Hence,jj(~1 , ••• ,~n)=O,j= 1,2, ... ,n, and this proves Lemma 1. D

Example 1. Consider the system of differential equations

dx 2 - =ax-bxy-ex dt '

dy - = -cy+dxy-jy2 dt

(2)

where a, b, c, d, e, and f are positive constants. This system (see Section 4.10) describes the population growth of two species x andy, where species y is dependent upon species x for its survival. Suppose that c I d > a I e. Prove that every solution x(t), y(t) of (2), with x(O) and y(O) > 0, approaches the equilibrium solution x = a I e, y = 0, as t approaches infinity. Solution. Our first step is to show that every solution x( t), y ( t) of (2) which starts in the first quadrant (x > 0, y > 0) at t = 0 must remain in the first quadrant for all future time. (If this were not so, then the model (2) could not correspond to reality.) To this end, recall from Section 1.5 that

ax0 x(l)= ,

ex0 + (a- ex0 )e-ar y(t)=O

is a solution of (2) for any choice of x0• The orbit of this solution is the point (0,0) for x0 = 0; the line 0 < x < al e for 0 < x0 < al e; the point (ale,O) for x0 =ale; and the line ale<x<oo for x0 >ale. Thus, the xaxis, for x ~ 0, is the union of four disjoint orbits of (2). Similarly, (see Exercise 14), the positive y-axis is a single orbit of (2). Thus, if a solution

429


y

a/b

II I

'<0, y<O

L-------------~--------~---------------- X ale c/d Figure 1

x(t), y(t) of (2) leaves the first quadrant, its orbit must cross another orbit, and this is precluded by the uniqueness of orbits (Property 1, Section 4.6).

Our next step is to divide the first quadrant into regions where dx I dt and dy I dt have fixed signs. This is accomplished by drawing the lines / 1 : a- by- ex =0, and /2 : - c + dx- fy =0, in the x-y plane. These lines divide the first quadrant into three regions I, II, and 111 as shown in Figure 1. (The lines /1 and /2 do not intersect in the first quadrant if c I d > a I e.) Now, observe that ex + by is less than a in region I, while ex + by is greater than a in regions II and 111. Consequently, dx I dt is positive in region I and negative in regions II and III. Similarly, dy I dt is negative in regions I and II and positive in region 111.

Next, we prove the following four simple Iemmas.

Lemma 3. Any solution x(t), y(t) of (2) which starts in region I at timet= t0

will remain in this region jor all future time t ;;;. t0 and ultimately approach the equilibrium solution x = a I e, y = 0.

PROOF. Suppose that a solution x(t), y(t) of (2) leaves region I at timet= t*. Then, .X( t*) = 0, since the only way a solution can leave region I is by crossing the line /1• Differentiating both sides of the first equation of (2) with respect to t and setting t = t* gives

d 2x(t*) dy(t*) dt2 = -bx(t*)~.

This quantity is positive. Hence, x( t) has a minimum at t = t*. But this is impossible, since x(t) is always increasing whenever x(t), y(t) is in region

430


I. Thus, any solution x(t), y(t) of (2) which starts in region I at timet= t0

will remain in region I for all future time t ~ t0 . This implies that x(t) is a monotonic increasing function of time, and y ( t) is · a monotonic decreasing function of time fort~ t0, with x(t)<ale andy(t)>O. Consequently, by Lemma I, both x(t) and y(t) have Iimits ~' 11 respectively, as t approaches infinity. Lemma 2 implies that (~, T/) is an equilibrium point of (2). Now, it is easily verified that the only equilibrium points of (2) in the region x ~ 0, y ~ 0 are x = 0, y = 0, and x = a I e, y = 0. Clearly, ~ cannot equal zero since x( t) is increasing in region I. Therefore, ~ = a I e and 11 = 0. 0

Lemma 4. Any solution x(t), y(t) of (2) which starts in region 111 at timet= t0 must leave this region at some later time.

PROOF. Suppose that a solution x(t), y(t) of (2) remains in region 111 for all time t ~ t0 • Then, x(t) is a monotonic decreasing function of time, and y(t) is a monotonic increasing function of time, fort~ t0• Moreover, x(t) is greater than c I d and y(t) is less than (dx(t0)- c)l f. Consequently, both x(t) and y(t) have Iimits ~' 11 respectively, as t approaches infinity. Lemma 2 implies that (~, 11) is an equilibrium point of (2). But (~, 11) cannot equal (0, 0) or ( a I e, 0) if x( t), y ( t) is in region III for t ~ t0 . This contradiction establishes Lemma 4. 0

Lemma 5. Any solution x(t), y(t) of (2) which starts in region II at time t = t0 and remains in region II for all future time t ~ t0 must approach the equilibrium solution x = a I e, y = 0.

PRooF. Suppose that a solution x(t), y(t) of (2) remains in region II for all timet~ t0• Then, both x(t) andy(t) are monotonic decreasing functions of timefort ~ t0 , with x(t)>O andy(t)>O. Consequently, by Lemma I, both x(t) and y(t) have Iimits ~' 11 respectively, as t approaches infinity. Lemma 2 implies that (~, T/) is an equilibrium point of (2). Now, (~, T/) cannot equal (0, 0). Therefore, ~ = a I e, 11 = 0. 0

Lemma 6. A solution x(t),y(t) of (2) cannot enter region III from region II.

PROOF. Suppose that a solution x(t), y(t) of (2) leaves region II at timet= t* and enters region III. Then, y( t*) = 0. Differentiating both sides of the second equation of (2) with respect to t and setting t = t* gives

d 2y(t*) dx(t*) 2 =4Y(t*)-d-.

dt t

This quantity is negative. Hence, y(t) has a maximum at t = t*. But this is impossible, sincey(t) is decreasing whenever x(t),y(t) is in region II. 0

Finally, observe that a solution x(t), y(t) of (2) which starts on 11 must immediately enter region I, and that a solution which starts on 12 must im-

431


mediately enter region II. It now follows immediately from Lemmas 3-6 that every solution x(t),y(t) of (2), with x(O)>O andy(O)>O, approaches the equilibrium solution x=aje,y=O as t approaches infinity.

Up to now, the solutions and orbits of the nonlinear equations that we have studied behaved very much like the solutions and orbits of linear equations. In actual fact, though, the situation is very different. The solutions and orbits of nonlinear equations, in general, exhibit a completely different behavior than the solutions and orbits of linear equations. A standard example is the system of equations

: = -y+x(1-x2-y2), ~ =x+y(1-x2-y2). (3)

Since the term x 2 + y 2 appears prominently in both equations, it suggests itself to introduce polar coordinates r,O, where x=rcosO,y=rsinO, and to rewrite (3) in terms of r and 0. To this end, we compute

d 2 dr dx dy dtr = 2r dt = 2x dt + 2y dt

=2(x2+ y2)-2(x2+y2)2 =2r2(1- r2).

Similarly, we compute

dy dx dO d Y 1 x dt - Y dt x 2 + y 2 - = -arctan- =- =---=I. dt dt x x2 1+(yjx)2 x2+y2

Consequently, the system of equations (3) is equivalent to the system of equations

dr 2 -=r(I-r) dt '

d() = 1 dt .

The general solution of (4) is easily seen to be ro

r( t) = I/2 '

[r6+(I-r6)e- 2']

where r0 = r(O) and 00 = 0(0). Hence,

ro x(t)= 112 cos(t+00 ),

[r6+(1-r6)e- 2']

ro . y(t)= 112 sm(t+00 ).

[ r6+ (1- r6)e- 21 ]

(4)

(5)

N ow, observe first that x = 0, y = 0 is the only equilibrium solution of (3). Second, observe that

x(t) =cos(t + 00 ), y(t) = sin(t + 00 )

432


y

Figure 2. The phase portrait of (3)

when r0 = 1. This solution is periodic with period 2'17, and its orbit is the unit circle x 2 + y 2 = 1. Finally, observe from (5) that r( t) approaches one as t approaches infinity, for r0 i=O. Hence, all the orbits of (3), with the exception of the equilibrium point x = 0, y = 0, spiral into the unit circle. This situation is depicted in Figure 2.

The system of equations (3) shows that the orbits of a nonlinear system of equations may spiral into a simple closed curve. This, of course, is not possible for linear systems. Moreover, it is often possible to prove that orbits of a nonlinear system spiral into a closed curve even when we cannot explicitly solve the system of equations, or even find its orbits. This is the content of the following celebrated theorem.

Theorem S. (Poincare-Bendixson.) Suppose that a solution x = x(t), y = y(t) of the system of differential equations

dx dy dt=j(x,y), dt=g(x,y) (6)

remains in a bounded region of the plane which contains no equi/ibrium points oj (6). Then, its orbit must spiral into a simple c/osed curoe, which is itself the orbit of a periodic solution of (6).

Example 2. Prove that the second-order differential equation

i+(z2 +2i2 -I)i+z=O (7)

has a nontrivial periodic solution.

433


Solution. First, we convert Equation (7) to a system of two first-order equations by setting x = z and y = i. Then,

dy dt = -x+(I-x2-2y2)y.

dx dt =y, (8)

Next, we try and find a bounded region R in the x-y plane, containing no equilibrium points of (8), and having the property that every solution x(t), y(t) of (8) which starts in R at time t = t0, remains there for allfuturetime t ~ t0• lt can be shown that a simply connected region such as a square or disc will never work. Therefore, we try and take R to be an annulus surrounding the origin. To this end, compute

!!_ ( x2+ y2) = x dx + y dy =(I- x2-2y2)y2 dt 2 dt dt ,

and observe that 1- x2- 2y2 is positive for x2 + y 2 < t and negative for x2 + y 2 > I. Hence, x2( t) + y 2( t) is increasing along any solution x( t), y ( t) of (8) when x2 + y 2 < t and decreasing when x2 + y 2 > I. This implies that any solution x(t),y(t) of (8) which starts in the annulus i<x2 +y2 <1 at time t = t0 will remain in this annulus for all future time t ~ t0• Now, this annulus contains no equilibrium points of (8). Consequently, by the Poincare-Bendixson Theorem, there exists at least one periodic solution x(t), y(t) of (8) lying entirely in this annulus, and then z = x(t) is a nontrivial periodic solution of (7).

EXERCISES

1. What Really Happened at the Paris Peace Talks

434

The original plan developed by Henry Kissinger and Le Duc Tho to settle the Vietnamese war is described below. It was agreed that 1 million South Vietnamese ants and 1 million North Vietnamese ants would be placed in the backyard of the Presidential palace in Paris and be allowed to fight it out for a long period of time. If the South Vietnamese ants destroyed nearly all the North Vietnamese ants, then South Vietnam would retain control of all of its land. If the North Vietnamese ants were victorious, then North Vietnam would take over all of South Vietnam. If they appeared to be fighting to a standoff, then South Vietnam would be partitioned according to the proportion of ants remaining. Now, the South Vietnamese ants, denoted by S, and the North Vietnamese ants, denoted by N, compete against each other according to the following differential equations:

dS 1 1 dt=WS- 20SXN

(•)

Note that these equations correspond to reality since the South Vietnamese ants multiply much more rapidly than the North Vietnamese ants, but the North Vietnamese ants are much better fighters.


The battle began at 10:00 sharp on the morning of May 19, 1972, and was supervised by a representative of Poland and a representative of Canada. At 2:43p.m. on the afternoon of May 21, the representative of Poland, being unhappy with the progress of the battle, slipped a bag of North Vietnamese ants into the backyard, but he was spotted by the eagle eyes of the representative of Canada. The South Vietnamese immediately claimed a foul and called off the agreement, thus setting the stage for the protracted talks that followed in Paris. The representative of Poland was hauled before a judge in Paris for sentencing. The judge, after making some remarks about the stupidity of the South Vietnamese, gave the Polish representative a very light sentence. Justify mathematically the judge's decision. Hint: (a) Show that the lines N = 2 and N + S = I divide the first quadrant into three

regions (see Figure 3) in which dS I dt and dN I dt have fixed signs. (b) Show that every solution S ( t), N ( t) of (*) which starts in either region I or

region 111 must eventually enter region II. (c) Show that every solution S(t),N(t) of (*) which starts in region II mustre

main there for all future time. (d) Conclude from (c) that S(t)~oo for all solutions S(t),N(t) of (*) with

S ( t0) and N ( t0) positive. Conclude too that N ( t) has a finite Iimit ( .;;; 2) as t~oo.

(e) To prove that N (t)~o. observe that there exists t0 such that dN I dt.;;; - N fort> t0• Conclude from this inequality that N(t)~O as t~oo.

N

m . . S< 0, N<O

2

1I S>O,N<O

1

Figure 3

2. Consider the system of differential equations

dx dy -=ax-bxy -=cy-dxy-ey2 dt ' dt

(*)

with alb>cle. Prove thaty(t)~O as t~oo, for every solu~on x(t),y(t) of (*) with x(t0) andy(t0) positive. Hint: Follow the outline in Exercise 1.

3. (a) Without computing the eigenvalues of the matrix

( -~ -D· 435


X1<0 .nz: Xa< 0 x,= 3x1

x,

JI[ x.< o x2>0

Figure 4

prove that every solution x(t) of

x=(-i I )x -3

approaches zero as t approaches infinity. Hint: (a) Show that the lines x2 = 3x1 and x1 =3x2 divide the x 1- x2 plane into four regions (see Figure 4) in which .X 1 and .X2 have fixed signs.

(b) Show that every solution x(t) which starts in either region I or II must remain there for all future time and ultimately approach the equilibrium solution x=O.

(c) Show that every solution x(t) which remains exclusively in region III or IV must ultimately approach the equilibrium solution x = 0.

A closed curve C is said to be a Iimit cycle of

x=f(x,y), .Y= g(x,y) (*)

if orbits of (*) spiral into it, or away from it. lt is stable if all orbits of (*) passing sufficiently close to it must ultimately spiral into it, and unstable otherwise. Find alllimit cycles of each of the following systems of differential equations. (Hint: Compute d(x 2 + y 2)/ dt. Observe too, that C must be the orbit of a periodic solution of (*) if it contains no equilibrium points of (*).)

x(x2 +y2 -2) 4 • .X= - y- ---====-

Yx2+y2

y(x2+y2-2) y =X- ---====--

Vx2+y2

436

4.9 Introduction to bifurcation theory

6. x= y+ x(x2+y2-l)(x2+ y 2-2) j= -x+y(x2+y2-1Xx2+y2-2)

8. (a) Show that the system

7. x=xy+xcos(x2+y2) j= -x2+ycos(x2+y2)

x=y+xf(r)/r, j= -x+yf(r)/r (*)

has Iimit cycles corresponding to the zeros of f(r). What is the direction of motion on these curves?

(b) Determine all Iimit cycles of (*) and discuss their stability if f(r)= (r-3)2(r2-5r+4).

Use the Poincare-Bendixson Theorem to prove the existence of a nontrivial periodic solution of each of the following differential equations.

11. (a) According to Green's theorem in the plane, if Cis a closed curve which is sufficiently "smooth," and if f and g are continuous and have continuous first partial derivatives, then

9i [f(x,y)dy- g(x,y)dx) = J J [fx(x,y)+ gy(x,y) J dxdy C R

where R is the region enclosed by C. Assurne that x(t), y(t) is a periodic solution of x = f(x,y), j = g(x,y), and Iet C be the orbit of this solution. Show that for this curve, the line integral above is zero.

(b) Suppose that fx + gY has the same sign throughout a simply connected region D in the x-y plane. Show that the system of equations x = f(x,y), j = g(x,y) can have no periodic solution which is entirely in D.

12. Show that the system of differential equations

x=x+y2+x3, j= -x+y+yx2

has no nontrivial periodic solution.

13. Show that the system of differential equations

has no nontrivial periodic solution which lies inside the circle x 2 + y 2 = 4.

14. (a) Show that x = 0, y = 1/;( t) is a solution of (2) for any function 1/;( t) satisfying ~=-ctJ;-ft/12.

(b) Choose 1/;( t0) > 0. Show that the orbit of x = 0, y = 1/;( t) (for all t for which 1/; exists) is the positive y axis.


Consider the system of equations

x=f(x,e) where

(1)

437


and e is a scalar. Intuitively speaking a bifurcation point of (1) is a value of e at which the solutions of (1) change their behavior. More precisely, we say that e= e0 is a bifurcation point of (I) if the phase portraits of (I) for e< e0

and e > e0 are different.

Remark. In the examples that follow we will appeal to our intuition in deciding whether two phase portraits are the same or are different. In more advanced courses we define two phase portraits to be the same, or topologically equivalent, if there exists a continuous transformation of the plane onto itself which maps one phase portrait onto the other.

Example 1. Find the bifurcation points of the system

x=Ax=(~ _nx Solution. The characteristic polynomial of the matrix A is

p(A.) = det(A- Al)

= det{ I -1 A. e ) -I-A.

= (A. -I)(A. +I)- e

=A.2 -(I+e).

(2)

The roots ofp(A.) are ±y'f+""E for e> -I, and ±/- e-I i for e< -I. This

12:

x1 = (1 +v'l+€) x2

Figure 1. Phaseportrait of (2) fort:> -1

438


x,

Figure 2. Phase portrait of (2) for e == - 1

implies that x = 0 is a saddle for e> -1, and a center for e< -1. We conclude, therefore, that e = - I is a bifurcation point of (2). lt is also clear that Eq. (2) has no other bifurcation points.

lt is instructive to see how the solutions of (2) change as e passes through the bifurcation value -1. For e > -1, the eigenvalues of A are

A1 =/I+e, A2 =-/I+e. lt is easily verified (see Exercise 10) that

xl=(l+v11+e)

is an eigenvector of A with eigenva1ue /I+E, whi1e

is an eigenvector with eigenva1ue -li+€. Hence, the phase portrait of (2) has the form shown in Figure 1. As e-> -1 from the 1eft, the lines /1 and /2

both approach the line x 1 = x2 . This 1ine is a 1ine of equilibrium points of (2) when e = -1, whi1e each line x 1 - x2 = c ( c'i' 0) is an orbit of (2) for e = - 1. The phase portrait of (2) for e = - 1 is given in Figure 2.

Example 2. Find the bifurcation points of the system

x = Ax = ( ~ = ~ )x. (3)

439


Solution. The characteristic polynomial of the matrix A is

p(A)=det(A-AI)=det( -eA

=A(l+A)+e

and the roots of p( A) are

-1 ) -1-A

Observe that A 1 is positive and A 2 is negative for e < 0. Hence, x = 0 is a saddle for e < 0. F or 0 < e < 114, both A 1 and A 2 are negative. Hence x = 0 is a stable node for 0 < e < 114. Both A 1 and A 2 are complex, with negative real part, for e> 114. Hence x = 0 is a stable focus for e> 114. Note that the phase portrait of (3) changes as e passes through 0 and 114. W e conclude, therefore, that e = 0 and e = 1 I 4 are bifurcation points of (3).

Example 3. Find the bifurcation points of the system of equations

dx 1 ---;]( = x2

dx2- 2 dt -xl-x2-e.

(4)

Solution. (i) We first find the equilibrium points of ( 4). Setting dx 11 dt = 0 gives x2 = 0, and then setting dx 2 ldt = 0 gives xf- e= 0, so that x 1 = ±1€, e > 0. Hence, ( 1€, 0) and (-1€, 0) are two equilibrium points of ( 4) for e > 0. The system ( 4) has no equilibrium points when e < 0. We conclude, therefore, that e= 0 is a bifurcation point of (4). (ü) We now analyze the behavior of the solutions of ( 4) near the equilibrium points ( ±.f€, 0) to determine whether this system has any additional bifurcation points. Setting

U = X 1 +-{e, V= X 2

gives

du dt =v

~~ =(u±{e)2 -v-e=±2{eu-v+u2 •

(5)

440


The system (5) can be written in the form

By Theorem 4, the phase portrait of (4) near the equilibrium solution

is determined by the phase portrait of the linearized system

To find the eigenvalues of A we compute

p(A) = det(A- Al)

( -A

= det ± 2{e

Hence, the eigenvalues of A when u = x 1 -Ii are

-I+JI+S{e -1-JI+S{e AI= 2 ' Az = 2 (6)

while the eigenvalues of A when u = x 1 + Ii are

-1+/t-s{e -l-~I-8{e AI= ___ 2 ___ ' Az = ___ 2 __ _ (7)

Observe from (6) that A1 > 0, while A2 < 0. Thus, the system (4) behaves like

a saddle near the equilibrium points ( ~ ) . On the other band, we see from

(7) that both A 1 and A 2 are negative for 0 < e < I I 64, and complex for e>1164. Consequently, the system (4) near the equilibrium solution

( -: ) behaves like a stable node for 0 < e < 1164, and a stable focus for

e> 1164. It can be shown that the phase portraits of a stable node and a stable focus are equivalent. Consequently, e = 1164 is not a bifurcation point of (4).

441


Another situation which is included in the context of bifurcation theory, and which is of much current research interest now, is when the system (1) has a certain number of equilibrium, or periodic, solutions for e = e0 , and a different number for e* e0 • Suppose, for example, that x(t) is an equilibrium, or periodic, solution of ( 1) for e = 0, and x 1( t ), x 2( t ), ... , x k( t) are equilibria, or periodic solutions of ( 1) for e * 0 which approach x( t) as e--> 0. In this case we say that the solutions x1(t),x2(t), ... ,xk(t) bifurcate from x(t). Weillustrate this situation with the following example.

Example 4. Find all equilibrium solutions of the system of equations

dx!- 2 2 dt-3ex 1 -3ex2 -x 1 -x2

dx dt = ex 1 - x 1x 2 =x1(e- x 2 ).

(8)

Solution. Let ( ;~) be an equilibrium solution of the system (8). The second

equation of (8) implies that x 1 = 0 or x 2 = e.

x 1 = 0. In this case, the first equation of (8) implies that

0 = 3ex2 + x~ = x 2(3e+ x 2 )

so that x 2 = 0 or x 2 =- 3e. Thus { ~) and { _03e) are two equilibrium

points of (8).

x 2 = e. In this case, the first equation of (8) imp1ies that

or

The solutions

of (9) are complex. Thus,

3e±}9e2 -16e2

x!= 2

and x 2 = { 0 ) -3e

(9)

are two equilibrium points of (8), for e * 0, which bifurcate from the single

equilibrium point x = { ~) when e = 0.

442

4.10 Predator-prey prob1ems

EXERCISES

Find the bifurcation points of each of the following systems of equations.

l.x={! 3. x={ ~2 ;)x 5. x =( ~ ~ E )x

2.x={! 4. x = ( ~

In each of Problems 6-8, show that more than one equilibrium solutions bifurcate from the equilibrium solution x = 0 when e = 0.

6. x1 = EX 1 - EX 2 - x? + xi Xz = EXz + x 1x 2

8. x1 = EX 2 + x 1x 2

Xz =- EX 1 + EXz + xf + xi

9. Consider the system of equations x1 = 3Ex 1-5Ex2 - xf + xi x2 = 2Ex 1 - EX 2 .

7. x]=Ex]-x?-x!x1 x2 = -2Ex1 +2Ex2 + x 1x 2 - xi

(a) Show that each point on the lines x 2 = x 1

points of ( *) for E = 0. and x 2 = - x 1 are equilibrium

(b) Show that

(XI) _ ( 0) ( X1) _ 7 ( 1) Xz - 0 and Xz -"JE 2 .

are the only equilibrium points of ( *) for E =!= 0.

10. Show that

are eigenvectors of the matrix { ~ _:. 1 ) with eigenvalues v"f+f and - v"f+f respectively.

4.10 Predator-prey problems; or why the percentage of sharks caught in the Mediterranean Sea rose dramatically during World War I

In the mid 1920's the ltalian biologist Umberto D'Ancona was studying the population variations of various species of fish that interact with each other. In the course of his research, he came across some data on per-

443


centages-of-total-catch of several species of fish that were brought into different Mediterranean ports in the years that spanned World War I. In particular, the data gave the percentage-of-total-catch of selachians, (sharks, skates, rays, etc.) which are not very desirable as food fish. The data for the port of Fiume, Italy, during the years 1914-1923 is given below.

1914 11.9%

1919 27.3%

1915 21.4%

1920 16.0%

1916 22.1%

1921 15.9%

1917 21.2%

1922 14.8%

1918 36.4%

1923 10.7%

D' Ancona was puzzled by the very large increase in the percentage of selachians during the period of the war. Obviously, he reasoned, the increase in the percentage of selachians was due to the greatly reduced Ievel of fishing during this period. But how does the intensity of fishing affect the fish populations? The answer to this question was of great concern to D' Ancona in bis research on the struggle for existence between competing species. It was also of concern to the fishing industry, since it would have obvious implications for the way fishing should be done.

Now, what distinguishes the selachians from the food fish is that the selachians are predators, while the food fish are their prey; the selachians depend on the food fish for their survival. At first, D' Ancona thought that this accounted forthelarge increase of selachians during the war. Since the Ievel of fishing was greatly reduced during this period, there were more prey available to the selachians, who therefore thrived and multiplied rapidly. However, this explanation does not hold any water since there were also more food fish during this period. D' Ancona's theory only shows that there are more selachians when the Ievel of fishing is reduced; it does not explain why a reduced Ievel of fishing is more beneficial to the predators than to their prey.

After exhausting all possible biological explanations of this phenomenon, D'Ancona turned to his colleague, the famous ltalian mathematician Vito Volterra. Hopefully, Volterra would formulate a mathematical model of the growth of the selachians and their prey, the food fish, and this model would provide the answer to D' Ancona's question. Volterra began bis analysis of this problern by separating all the fish into the prey population x(t) and the predator populationy(t). Then, he reasoned that the food fish do not compete very intensively among themselves for their food supply since this is very abundant, and the fish population is not very dense. Hence, in the absence of the selachians, the food fish would grow according to the Malthusian law of population growth x = ax, for some positive constant a. Next, reasoned Volterra, the nurober of contacts per unit time between predators and prey is bxy, for some positive constant b. Hence, x = ax- bxy. Similarly, Volterra concluded that the predators have a natural rate of decrease - cy proportional to their present number, and

444

4.10 Predator-prey problems

that they also increase at a rate dxy proportional to their present number y and their food supply x. Thus,

dx -=ax-bxy dt '

dy dt = - cy + dxy. (1)

The system of equations (1) governs the interaction of the selachians and food fish in the absence of fishing. W e will carefully analyze this system and derive several interesting properties of its solutions. Then, we will include the effect of fishing in our model, and show why a reduced level of fishing is more beneficial to the selachians than to the food fish. In fact, we will derive the surprising result that a reduced level of fishing is actually harmful to the food fish.

Observe first that (1) has two equilibrium solutions x(t)=O,y(t)=O and x(t)= cl d, y(t) = al b. The first equilibrium solution, of course, is of no interest to us. This systemalso has the family of solutions x(t) = x0ea1, y(t) = 0 and x(t)=O,y(t)=y0e-c1• Thus, both the x andy axes are orbits of (1). This implies that every solution x(t), y(t) of (1) which starts in the first quadrant x > 0, y > 0 at time t = t0 will remain there for all future time t ~ lo.

The orbits of (1), for x,y;FO are the solution curves of the first-order equation

dy -cy+dxy y(-c+dx)

dx = ax - bxy x ( a - by) · (2)

This equation is separable, since we can write it in the form

a-by dy -c+dx y dx x

Consequently, a lny- by + clnx- dx = k 1, for some constant k 1• Taking exponentials of both sides of this equation gives

(3)

for some constant K. Thus, the orbits of (1) are the family of curves defined by (3), and these curves are closed as we now show.

Lemma 1. Equation (3) defines a family of closed curves for x, y > 0.

PROOF. Our first step is to determine the beha vior of the functions f (y) =

y a I eby and g(x) = xc I edx for x and y positive. To this end, observe that f(O) = 0, f( oo) = 0, and f(y) is positive for y > 0. Computing

aya-l_bya ya-l(a-by) f' (y) = by = by '

e e

we see thatf(y) has a single critical point aty=alb. Consequently,j(y) achieves its maximum value My=(albtlea aty=alb, and the graph of

445


f(y)

--~------~----~==~y alb

(a)

g(x)

--~------~--------~x c/d (b)

Figure 1. (a) Graph off(y)=yae-hY; (b) Graph of g(x)=xce-dx

j(y) has the form described in Figure la. Similarly, g(x) achieves its maximum value Mx =(cl dY I ec at x = cl d, and the graph of g(x) has the form described in Figure 1 b.

From the preceding analysis, we conclude that Equation (3) has no solution x,y>O for K>MxMy, and the single solution x=cld,y=alb for K =Mx MY. Thus, we need only consider the case K = A.MY, where A. is a positive number less than Mx. Observe first that the equation xcledx=A. has one solution x = xm < c I d, and one solution x = xM > c I d. Hence, the equation

has no solutiony when x is less than xm or greater than xM. It has the single solutiony = al b when x =xm or xM, and it has two solutionsy 1(x) and h(x) for each x between xm and xM. The smaller solution y 1(x) is always less than al b, while the larger solution h(x) is always greater than a/ b. As x approaches either xm or xM, bothy 1(x) andyix) approach al b. Consequently, the curves defined by (3) are closed for x and y positive, and have the form described in Figure 2. Moreover, none of these closed curves (with the exception of x = c I d, y = a I b) contain any equilibrium points of (1). Therefore, all solutions x(t), y(t) of (1), with x(O) and y(O) positive, are periodic functions of time. That is to say, each solution x(t), y(t) of (1), with x(O) andy(O) positive, has the property that x(t+ T)=x(t) and y(t + T) = y(t) for some positive T. 0

Now, the data of D'Ancona is really an average over each one year period of the proportion of predators. Thus, in order to compare this data with the predictions of (1), we must compute the "average values" of x(t) and y(t), for any solution x(t), y(t) of (1). Remarkably, we can find these average values even though we cannot compute x(t) andy(t) exactly. This is the content of Lemma 2.

446


y

o/b •

L-------~------------~~-------------J---X xm c/d

Figure 2. Orbits of (1) for x,y positive

Lemma 2. Let x( t), y ( t) be a periodic solution of (I), with period T > 0. Define the average values of x and y as

I (T x= T Jo x(t)dt,

I (T y = T Jo y(t)dt.

Then, x=cld andy=alb. In other words, the average values of x(t) and y(t) are the equilibrium values.

PRooF. Dividing both sides of the first equation of (1) by x gives .X I x = a-by, so that

I (T i(t) I (T T }

0 x(t)dt= T)o [a-by(t)]dt.

Now, JoTx(t)lx(t)dt=Inx(T)-Inx(O), and this equals zero since x(T)=

x(O). Consequently,

l. (Tby(t)dt= l. (Tadt=a, T Jo T Jo

so that y = a I b. Similarly, by dividing both sides of the second equation of (I) by Ty ( t) and integrating from 0 to T, we obtain that .X= c I d. D

We are now ready to include the effects of fishing in our model. Observe that fishing decreases the population of food fish at a rate u(t), and decreases the population of selachians at a rate ey(t). The constant e reflects the intensity of fishing; i.e., the number of boats at sea and the number of nets in the water. Thus, the true state of affairs is described by the

447


modified system of differential equations

dx = ax- b.xy- ex= (a- e)x- b.xy dt dy dt = - cy + dxy - ey = - ( c + e) y + dxy.

(4)

This system is exactly the same as (1) (for a- e > 0), with a replaced by a-e, and c replaced by c+e. Hence, the average values of x(t) andy(t) are now

- c+e x=-d , - a-e y=-b-. (5)

Consequently, a moderate amount of fishing (e < a) actually increases the number of food fish, on the average, and decreases the number of selachians. Conversely, a reduced Ievel of fishing increases the number of selachians, on the average, and decreases the number of food fish. This remarkable result, which is known as Volterra's principle, explains the data of D' Ancona, and completely solves our problem.

Volterra's principle has spectacular applications to insecticide treatments, which destroy both insect predators and their insect prey. lt implies that the application of insecticides will actually increase the population of those insects which are kept in control by other predatory insects. A remarkable confirmation comes from the cottony cushion scale insect (Icerya purchasi), which, when accidentally introduced from Australia in 1868, threatened to destroy the American citrus industry. Thereupon, its natural Australian predator, a ladybird beetle (Novius Cardinalis) was introduced, and the beetles reduced the scale insects to a low Ievel. When DDT was discovered to kill scale insects, it was applied by the orchardists in the hope of further reducing the scale insects. However, ·in agreement with Volterra's principle, the effect was an increase of the scale insect!

Oddly enough, many ecologists and biologists refused to accept Volterra's model as accurate. They pointed to the fact that the oscillatory behavior predicted by Volterra's model is not observed in most predator-prey systems. Rather, most predator-prey systems tend to equilibrium states as time evolves. Our answer to these critics is that the system of differential equations (I) is not intended as a model of the general predator-prey interaction. This is because the food fish and selachians do not compete intensively among themselves for their available resources. A more general model of predator-prey interactions is the system of differential equations

x=ax-bxy-ex2, y=-cy+d.xy-jy2• (6)

Here, the term ex2 reflects the internal competition of the prey x for their limited external resources, and the termjy2 reflects the competition among the predators for the limited number of prey. The solutions of (6) are not, in general, periodic. Indeed, we have already shown in Example 1 of Sec-

448


tion 4.8 that all solutions x(t), y(t) of (6), with x(O) and y(O) positive, ultimately approach the equilibrium solution x = a I e, y = 0 if c I d is greater than al e. In this situation, the predators die out, since their available food supply is inadequate for their needs.

Surprisingly, some ecologists and biologists even refuse to accept the moregenerat model (6) as accurate. As a counterexample, they cite the experiments of the mathematical biologist G. F. Gause. In these experiments, the population was composed of two species of protozoa, one of which, Didinium nasatum, feeds on the other, Paramecium caudatum. In all of Gause's experiments, the Didinium quickly destroyed the Paramecium and then died of starvation. This situation cannot be modeled by the system of equations (6), since no solution of (6) with x(O)y(O)*O can reach x=O or y = 0 in finite time.

Our answer to these critics is that the Didinium are a special, and atypical type of predator. On the one hand, they are ferocious attackers andrequire a tremendous amount of food; a Didinium demands a fresh Paramecium every three hours. On the other hand, the Didinium don't perish from an insufficient supply of Paramecium. They continue to multiply, but give birth to smaller offspring. Thus, the system of equations (6) does not accurately model the interaction of Paramecium and Didinium. A better model, in this case, is the system of differential equations

dx ,c -=ax-bvxy dt '

dy = { dYx y, dt -cy,

x*O x=O

(7)

lt can be shown (see Exercise 6) that every solution x(t), y(t) of (7) with x(O) andy(O) positive reaches x=O in finite time. This does not contradict the existence-uniqueness theorem, since the function

g(x,y)= { dYx y, x*O -cy, x=O

does not have a partial derivative with respect to x or y, at x = 0. Finally, we mention that there are several predator-prey interactions in

nature which cannot be modeled by any system of ordinary differential equations. These situations occur when the prey are provided with a refuge that is inaccessible to the predators. In these situations, it is impossible to make any definitive Statements about the future number of predators and prey, since we cannot predict how many prey will be stupid enough to leave their refuge. In other words, this process is now random, rather than deterministic, and therefore cannot be modeled by a system of ordinary differential equations. This was verified directly in a famous experiment of Gause. He placed five Paramecium and three Didinium in each of thirty identical test tubes, and provided the Paramecium with a refuge from the Didinium. Two days later, he found the predators dead in four tubes, and a mixed population containing from two to thirty-eight Paramecium in the remaining twenty-six tubes.

449


Reference Volterra, V: "Leyons sur la theorie mathematique de la lutte pour la vie." Paris,

1931.

EXERCISES

1. Findall biologically realistic equilibrium points of (6) and determine their stability.

2. We showed in Section 4.8 thaty(t) ultimately approaches zero for all solutions x(t),y(t) of (6), if c/d>aje. Show that there exist solutions x(t),y(t) of (6) for whichy(t) increases at first to a maximum value, and then decreases to zero. (To an observer who sees only the predators without noticing the prey, such a case of a population passing through a maximum to total extinction would be very difficult to explain.)

3. In many instances, it is the adult members of the prey who are chiefly attacked by the predators, while the young members are better protected, either by their smaller size, or by their living in a different station. Let x 1 be the number of adult prey, x2 the number of young prey, and y the number of predators. Then,

.X1= -a1x 1+a2x 2-bx1y

x2= nxl- (al + a2)x2

y= -cy+dx1y

where a2x2 represents the number of young (per unit time) growing into adults, and n represents the birth rate proportional to the number of adults. Find all equilibrium solutions of this system.

4. There are several situations in nature where species 1 preys on species 2 which in turn preys on species 3. One case of this kind of population is the Island of Komodo in Malaya which is inhabited by giant carnivorous reptiles, and by mammals-their food-which feed on the rieb vegetation of the island. We assume that the reptiles have no direct influence on the vegetation, and that only the plants compete among themselves for their available resources. A system of differential equations governing this interaction is

.XI= -alxl- b12x1x2+c13x1x3

X2 =- Q2X2 + b21XIX2

X3= a3X3 -a4x~- C31X1X3

Find all equilibrium solutions of this system.

5. Consider a predator-prey system where the predator has alternate means of support. This system can be modelled by the differential equations

.X1 = a 1x 1 ( ß1 - x 1) + y 1x 1x 2

X2= a2x2( ß2- X2)- 'Y2X1X2

where x 1(t) and x2(t) are the predators and prey populations, respectively, at timet.

450

4.11 The principle of competitive exclusion in population biology

(a) Show that the change of coordinates ß;Y;(t)=x;(t/a;ß;) reduces this system of equations to

YJ = Y1 (1-YJ)+aiYJY2• h= Y2(1-Y2)- a2Y1Y2

where a 1 = Y1 ß2/ 01.1 ß1 and a2 = Y2 ßd 01.2 ß2· (b) What are the stable equilibrium populations when (i) 0 < a2 < I, (ii) a2 > l? ( c) It is observed that a 1 = 3a2 ( a2 is a measure of the aggressiveness of the pre

dator). What is the value of a2 if the predator's instinct is to maximize its stable equilibrium population?

6. (a) Let x(t) be a solution of .X= ax- MVx, with M > a Vx{tJ . Show that

aVx =M-(M-aVx{iJ )ea(t-to)/2.

(b) Conclude from (a) that x(t) approaches zeroinfinite time.

(c) Let x(t), y(t) be a solution of (7), with by(t0) > a Vx(tJ . Show that x(t) reaches zero in finite time. Hint: Observe that y(t) is increasing fort> t0•

( d) It can be shown that by ( t) will eventually exceed a -v-;(i) for every solution x(t), y(t) of (7) with x(t0) and y(t0) positive. Conclude, therefore, that all solutions x(t), y(t) of (7) achieve x=O in finite time.

4.11 The principle of competitive exclusion in population biology

lt is often observed, in nature, that the struggle for existence between two similar species competing for the same limited food supply and living space nearly always ends in the complete extinction of one of the species. This phenomenon is known as the "principle of competitive exclusion." lt was first enunciated, in a slightly different form, by Darwin in 1859. In his paper 'The origin of species by natural selection' he writes: "As the species of the same genus usually have, though by no means invariably, much similarity in habits and constitutions and always in structure, the struggle will generally be more severe between them, if they come into competition with each other, than between the species of distinct genera."

There is a very interesting biological explanation of the principle of competitive exclusion. The cornerstone of this theory is the idea of a "niche." A niche indicates what place a given species occupies in a community; i.e., what are its habits, food and mode of life. It has been observed that as a result of competition two similar species rarely occupy the same niche. Rather, each species takes possession of those kinds of food and modes of life in which it has an advantage over its competitor. If the two species tend to occupy the same niche then the struggle for existence between them will be very intense and result in the extinction of the weaker species.

An excellent illustration of this theory is the colony of terns inhabiting the island of Jorilgatch in the Black Sea. This colony consists of four different species of terns: sandwich-tern, common-tern, blackbeak-tern, and lit-

451


tle-tern. These four species band together to chase away predators from the colony. However, there is a sharp difference between them as regards the procuring of food. The sandwich-tern flies far out into the open sea to hunt certain species, while the blackbeak-tern feeds exclusively on land. On the other hand, common-tern and Iittle-tern catch fish close to the shore. They sight the fish while flying and dive into the water after them. The Iittle-tern seizes his fish in shallow swampy places, whereas the common-tern hunts somewhat further from shore. In this manner, these four similar species of tern living side by side upon a single small island differ sharply in all their modes of feeding and procuring food. Each has a niche in which it has a distinct advantage over its competitors.

In this section we present a rigorous mathematical proof of the law of competitive exclusion. This will be accomplished by deriving a system of differential equations which govern the interaction between two similar species, and then showing that every solution of the system approaches an equilibrium state in which one of the species is extinct.

In constructing a mathematical model of the struggle for existence between two competing species, it is instructive to Iook again at the logistic law of population growth

(1)

This equation governs the growth of the population N ( t) of a single species whose members compete among themselves for a limited amount of food and living space. Recall (see Section 1.5) that N (t) approaches the limiting population K = a I b, as t approaches infinity. This limiting population can be thought of as the maximum population of the species which the microcosm can support. In terms of K, the logistic law (I) can be rewritten in the form

Equation (2) has the following interesting interpretation. When the population N is very low, it grows according to the Malthusian law dN I dt =aN. The term aN is called the "biotic potential" of the species. It is the potential rate of increase of the species under ideal conditions, and it is realized if there are no restrictions on food and living space, and if the individual members of the species do not excrete any toxic waste products. As the population increases though, the biotic potential is reduced by the factor (K- N)l K, which is the relative number of still vacant places in the microcosm. Ecologists call this factor the environmental resistance to growth.

Now, Iet N 1(t) and N 2(t) be the population at timet of species 1 and 2 respectively. Further, Iet K 1 and K 2 be the maximum population of species 1 and 2 which the microcosm can support, and 1et a1N 1 and a2N 2 be the biotic potentials of species 1 and 2. Then, N 1(t) and Nit) satisfy the sys-

452

4.11 The princip1e of competitive exclusion in popu1ation bio1ogy

(3)

where m2 is the total number of places of the first species which are taken up by members of the second species, and m 1 is the total number of places of the second species which are taken up by members of the first species. At first glance it would appear that m2 = N2 and m 1 = N 1• However, this is not generally the case, for it is highly unlikely that two species utilize the environment in identical ways. Equal numbers of individuals'of species I and 2 do not, on the average, consume equai quantities of food, take up equal amounts of living space and excrete equal amounts of waste products of the same chemical composition. In general, we must set m2 = a.N1

and m 1 = ßN1, for some constants a. and ß. The constants a. and ß indicate the degree of influence of one species upon the other. If the interests of the two species do not clash, and they occupy separate niches, then both a. and ß are zero. If the two species lay claim to the same niche and are very similar, then a. and ß are very close to one. On the other hand, if one of the species, say species 2, utilizes the environment very unproductively; i.e., it consumes a great deal of food or excretes very poisonous waste products, then one individual of species 2 takes up the place of many individuals of species l. In this case, then, the coefficient a. is very large.

We restriet ourselves now to the case where the two species are nearly identical, and lay claim to the same niche. Then, a. = ß = 1, and N 1(t) and Nit) satisfy the system of differential equations

(4)

In this instance, we expect the struggle for existence between species I and 2 to be very intense, and to result in the extinction of one of the species. This is indeed the case as we now show.

Theorem 6 (Principle of competitive exclusion). Suppose that K 1 is greater than K2• Then, every solution N 1(t), Nit) of (4) approaches the equilibrium solution N 1 = K1, N2 = 0 as t approaches infinity. In other words, if species I and 2 are very nearly identical, and the microcosm can support more members of species 1 than of species 2, then species 2 will ultimately become extinct.

Our first step in proving Theorem 6 is to show that N 1(t) and Nit) can never become negative. To this end, recall from Section 1.5 that

453


is a solution of (4) for any choice of NI(O). The orbit of this solution in the NI-N2 plane is the point (0,0) for NI(O)=O; the line 0< NI< KI, N2=0 for O<NI(O)<KI; the point (KI,O) for NI(O)=KI; and the line KI <NI <oo, N2 =0 for NI(O)> KI. Thus, the NI axis, for NI;;;. 0, is the union of four distinct orbits. Similarly, the N 2 axis, for N 2 ;;;. 0, is the union of four distinct orbits of (4). This implies that all solutions NI(t), N2(t) of (4) which start in the first quadrant (NI >0,N2 >0) of the NI-N2 plane must remain there for all future time.

Our second step in proving Theorem 6 is to split the first quadrant into regions in which both dNI/ dt and dNd dt have fixed signs. This is accomplished in the following manner. Let /I and /2 be the lines KI- NI- N 2 = 0 and K 2 - NI- N2 = 0, respectively. Observe that dNI/ dt is negative if (NI,N2) lies above /I, and positive if (NI,N2) lies below /I. Similarly, dN2/dt is negative if (NI,N2) lies above /2> and positive if (NI,N~ lies below /2. Thus, the two parallellines /I and /2 split the first quadrant of the NI-N2 plane into three regions (see Figure I) in which both dNI/ dt and dNd dt have fixed signs. Both NI(t) and Nit) increase with time (along any solution of (4)) in region I; NI(t) increases, and Nit) decreases, with time in region II; and both NI(t) and Nit) decrease with time in region III.

lii

I ~,> 0 N2>o

~~------~~--------~------- N, K,

Figure l

Lemma l. Any solution NI(t), Nit) of (4) which starts in region I at t = t0

must leave this region at some later time.

PROOF. Suppose that a solution NI(t), Nit) of (4) remains in region I for all time t;;;. t0 • This implies that both NI(t) and Nit) are monotonic increasing functions of time for t;;;. t0, with NI(t) and Nit) less than K2•

Consequently, by Lemma I of Section 4.8, both NI(t) and N2(t) have Iimits

454

4.11 The princip1e of competitive exclusion in population bio1ogy

~. 'IJ respectively, as t approaches infinity. Lemma 2 of Section 4.8 implies that (~,'IJ) is an equilibrium point of (4). Now, the only equilibrium points of (4) are (0,0), (K1,0), and (O,K2), and (~,'IJ) obviously cannot equal any of these three points. We conclude therefore, that any solution N 1(t), N2(t) of (4) which starts in region I must leave this region at a later time. 0

Lemma 2. Any solution N 1(t), N2(t) of (4) which starts in region II at time t = t0 will remain in this region for all future time t ;> t0, and ultimately approach the equilibrium solution N 1 =K1, N2 =0.

PR.ooF. Suppose that a solution N1(t), N2(t) of (4) leaves region II at time t = t*. Then, either N 1(t*) or Nit*) is zero, since the only way a solution of (4) can leave region II is by crossing /1 or /2• Assurne that N1(t*)=O. Differentiating both sides of the first equation of (4) with respect tot and setting t = t* gives

-a1N 1 (t*) dN2 (t*)

K 1 dt

This quantity is positive. Hence, N 1(t) has a minimum at t=t*. Butthis is impossible, since N 1(t) is increasing whenever a solution N 1(t), N2(t) of (4) is in region II. Similarly, if Nit*)=O, then

d 2N 2 (t*) -a2N 2 (t*) dN1 (t*)

dt

This quantity is negative, implying that N2(t) has a maximum at t= t*. But this is impossible, since N2(t) is decreasing whenever a solution N 1(t), N2(t) of (4) is in region II.

The previous argument shows that any solution N 1(t), N2(t) of (4) which starts in region II at time t = t0 will remain in region II for all future time t > t0• This implies that N 1(t) is monotonic increasing and N2(t) is monotonic decreasing for t;>t0, with N 1(t)<K1 and N2(t)>K2• Consequently, by Lemma 1 of Section 4.8, both N 1(t) and N2(t) have limits ~,'IJ respectively, as t approaches infinity. Lemma 2 of Section 4.8 implies that (~, 'IJ) is an equilibrium point of (4). Now, (~, 'IJ) obviously cannot equal (0, 0) or (O,K2). Consequently, (~,1J)=(K1 ,0), and this proves Lemma 2. 0

Lemma 3. Any so/ution N 1(t), N2(t) of (4) which starts in region 111 at time t = t0 and remains there for all future time must approach the equilibrium solution N 1(t)=K1, N2(t)=O as t approaches infinity.

PR.ooF. If a solution N1(t), N2(t) of (4) remains in region 111 fort> t0, then both N 1(t) and N2(t) are monotonic decreasing functions of timefort ;> tO> with N 1(t)>O and N2(t)>O. Consequently, by Lemma 1 of Section 4.8, both N 1(t) and N2(t) have limits ~,'IJ respectively, as t approaches infinity. Lemma 2 of Section 4.8 implies that (~,'IJ) is an equilibrium point of (4). Now, (~,'IJ) obviously cannot equal (0,0) or (O,KJ. Consequently, (~,1J)= (K1,0). 0

455


PRooF OF THEOREM 6. Lemmas I and 2 above state that every solution N1(t), N2(t) of (4) which starts in regions I or II at time t=t0 must approach the equilibrium solution N1 = K1, N2 =0 as t approaches infinity. Similarly, Lemma 3 shows that every solution N1(t), N2(t) of (4) which starts in region 111 at time t = t0 and remains there for all future time must also approach the equilibrium solution N1 = K1, N2 = 0. Next, observe that any solution N 1(t), Nit) of (4) which starts on /1 or /2 must immediately afterwards enter region II. Finally, if a solution N1(t),N2(t) of (4) leaves region 111, then it must cross the line /1 and immediately afterwards enter region II. Lemma 2 then forces this solution to approach the equilibrium solution N 1 = K1, N2 = 0. D

Theorem 6 deals with the case of identical species; i.e., a = ß = 1. By a similar analysis (see Exercises 4-6) we can predict the outcome of the struggle for existence for all values of a and ß.

Reference Gause, G. F., 'The Struggle for Existence,' Dover Publications, New York, 1964.

EXERCISES

1. Rewrite the system of equations (4) in the form

K 1 dN1 K2 dN2 ----=K1-N1-N2, ----=K2-N1-N2• a 1N 1 dt a2N 2 dt

Then, subtract these two equations and integrate to obtain directly that N2(t) approaches zero for all solutions N1(t), N2(t) of (4) with N1(t0)>0.

2. The system of differential equations

dN1 dt =N1 [ -al +c1(l-b1N1 -b2N2 )]

dN2 dt =N2 [ -a2 + c2 (l-b1N 1 -b2N2 )]

(*)

is a model of two species competing for the same limited resource. Suppose that c1 > a1 and c2 > a2• Deduce from Theorem 6 that N1(t) ultimately approaches zero if a1c2 >a2c1, and N2(t) ultimately approaches zero if a1c2 <a2c1•

3. In 1926, Volterra presented the following model of two species competing for the same limited food supply:

dN1 dt =[bJ-A1(h1N1+h2N2)]N1

dN2 dt =[b2-X2(h1N1 +h2N2 )]N2•

Suppose that bJ/A1 > bdX2• (The coefficient b;/A; is called the susceptibility of species i to food shortages.) Prove that species 2 will ultimately become extinct if N1(t0)>0.

456

4.11 The princip1e of competitive exclusion in popu1ation bio1ogy

Problems 4-6 are concerned with the system of equations

dN1 a 1N 1 dN2 a2N2 dt=K;""(KI-NI-aN2), dt= K2 (K2-N2-ßNI). (*)

4. (a) Assurne that K1/ a > K2 and Kd ß < K1• Show that N2(t) approaches zero as t approaches infinity for every solution N 1(t), N 2(t) of (*) with N 1(t0) > 0.

(b) Assurne that Kif a < K2 and K2/ ß > K1• Show that N 1(t) approaches zero as t approaches infinity for every solution N 1(t), N2(t) of (*) with N 1Nit0)>0. Hint: Draw the lines /1 :N1 +aN2 = K1 and /2 :N2 + ßN1 =K2, and follow the proof of Theorem 6.

5. Assurne that K1/ a > K2 and K2/ ß > K1• Prove that all solutions N 1(t), N 2(t) of (*), with both N 1(t0) and Nit0) positive, ultimately approach the equilibrium solution

Hint: (a) Draw the lines /1 :N1 + aN2 = K1 and /2 :N2 + ßN1 =K2• The two 1ines divide

the first quadrant into four regions (see Figure 2) in which both N1 and N2

have fixed signs.

I N?O N>O 2

~----------~--~------~--------------NI K,

Figure 2

(b) Show that all solutions N 1(t), N 2(t) of (*) which start in either region II or III must remain in these regions and ultimately approach the equilibrium solution N 1 =NP, N2 =Nf.

(c) Show that all solutions N 1(t), N2(t) of (*) which remain exclusively in region I or region IV for all time t ;;. t0 must ultimately approach the equilibrium solution N 1 =NP, N2 = Nr.

457


K/a

Figure 3

6. Assurne that K1/ a < K2 and K2/ ß < K1•

(a) Show that the equilibrium solution N1 =0, N2 =0 of (*) is unstable. (b) Show that the equilibrium solutions N1 =K~> N2 =0 and N1=0, N 2=K2 of

(*) are asymptotically stable. (c) Show that the equilibrium solution N1 = Nf, N 2 = Nf (see Exercise 5) of (*)

is a saddle point. (This calculation is very cumbersome.) (d) It is not too difficult to see that the phase portrait of (*) must have the form

described in Figure 3.

4.12 The Threshold Theorem of epidemiology

Consider the situation where a small group of people having an infectious disease is inserted into a large population which is capable of catching the disease. What happens as time evolves? Will the disease die out rapidly, or will an epidemic occur? How many people will ultimately catch the disease? To answer these questions we will derive a system of differential equations which govern the spread of an infectious disease within a population, and analyze the behavior of its solutions. This approach will also lead us to the famous Threshold Theorem of epidemiology which states that an epidemic will occur only if the number of people who are susceptible to the disease exceeds a certain threshold value.

W e begin with the assumptions that the disease under consideration confers permanent immunity upon any individual who has completely recovered from it, and that it has a negligibly short incubation period. This latter assumption implies that an individual who contracts the disease becomes infective immediately afterwards. In this case we can divide the population into three classes of individuals: the infective class (/), the susceptible class (S) and the removed class (R). The infective class consists of those individuals who are capable of transmitting the disease to others.

458


The susceptible dass consists of those individuals who are not infective, but who are capable of catching the disease and becoming infective. The removed dass consists of those individuals who have had the disease and are dead, or have recovered and are permanently immune, or are isolated until recovery and permanent immunity occur.

The spread of the disease is presumed to be governed by the following rules.

Rule 1: The population remains at a fixed Ievel N in the time interval under consideration. This means, of course, that we neglect births, deaths from causes unrelated to the disease under consideration, immigration and emigration.

Rule 2: The rate of change of the susceptible population is proportional to the product of the number of members of ( S) and the number of members of (/).

Rule 3: Individuals are removed from the infectious dass (/) at a rate proportional to the size of (I).

Let S ( t),/ ( t), and R ( t) denote the number of individuals in classes ( S), (I), and (R), respectively, at timet. lt follows immediately from Rules l-3 that S (t),l ( t), R (t) satisfies the system of differential equations

dS = -rSI dt di dt =rSI-yi (1)

dR =yi dt

for some positive constants r and y. The proportionality constant r is called the infection rate, and the proportionality constant y is called the removal rate.

The first two equations of (1) do not depend on R. Thus, we need only consider the system of equations

dS -= -rSI dt '

di - =rSI-yi dt

(2)

for the two unknown functions S (t) and I (t). Once S (t) and I (t) are known, we can solve for R (t) from the third equation of (1). Alternately, observe that d(S+I+R)/dt=O. Thus,

S (t)+ I (t) + R (t) =constant= N so that R ( t) = N- S ( t)- I ( t).

The orbits of (2) are the solution curves of the first-order equation

di rSI-yi y dS = - rSI = - l + rS . (J)

Integrating this differential equation gives s

I(S)= I0 + S0 - S+plns, 0

(4)

459


I

(So,Io)

~--~--~---L------~------------5 p

Figure 1. The orbits of (2)

where S0 and I0 are the number of susceptibles and infectives at the initial timet= t0 , and p = y Ir. To analyze the behavior of the curves (4), we compute I'(S)= -1 +PIS. The quantity -1 +PIS is negative for S > p, and positive for S < p. Hence, I (S) is an increasing function of S for S < p, and a decreasing function of S for S > p.

N ext, observe that I (0) = - oo and I ( S0) = I 0 > 0. Consequently, there exists a unique point S~, with 0< S~ < S0, suchthat I(S~)=O, and I(S) > 0 for S ~ < S ,.;; S 0. The point ( S ~, 0) is an equilibrium point of (2) since both dS I dt and di I dt vanish when I= 0. Thus, the orbits of (2), for t0 ..; t < oo, have the form described in Figure I.

Let us see what all this implies about the spread of the disease within the population. Astruns from t0 to oo, the point (S(t),/(t)) travels along the curve (4), and it moves along the curve in the direction of decreasing S, since S (t) decreases monotonically with time. Consequently, if S0 is less than p, then I ( t) decreases monotonically to zero, and S ( t) decreases monotonically toS~. Thus, if a small group of infectives I0 is inserted into a group of susceptibles S0, with S0 < p, then the disease will die out rapid1y. On the other hand, if S0 is greater than p, then I ( t) increases as S ( t) decreases to p, and it achieves a maximum value when S = p. lt only starts decreasing when the number of susceptibles falls below the threshold value p. From these results we may draw the following conclusions.

Conclusion 1: An epidemic will occur only if the number of susceptibles in a population exceeds the threshold value p = y Ir.

Conclusion 2: The spread of the disease does not stop for lack of a susceptible population; it stops only for lack of infectives. In particular, some individuals will escape the disease altogether.

Conclusion 1 corresponds to the general observation that epidemics tend to build up more rapidly when the density of susceptibles is high due to overcrowding, and the removal rate is low because of ignorance, inadequate isolation and inadequate medical care. On the other hand, outbreaks tend to be of only limited extent when good social conditions entaillower

460


densities of susceptibles, and when removal rates are high because of good public health vigilance and control.

If the number of susceptibles S0 is initially greater than, but close to, the threshold value p, then we can estimate the number of individuals who ultimately contract the disease. Specifically, if S0 - p is small compared to p, then the number of individuals who ultimately contract the disease is approximately 2( S0 - p ). This is the famous Threshold Theorem of epidemiology, which was first proven in 1927 by the mathematical biologists Kermack and McKendrick.

Theorem 7 (Threshold Theorem of epidemiology). Let S0 = p + v and assume that v / p is very small compared to one. Assurne moreover, that the number of initial infectives / 0 is very small. Then, the number of individuals who ultimately contract the disease is 2v. In other words, the Ievel of susceptibles is reduced to a point as far below the Ihreshold as it origina/ly was above it.

PROOF. Letting t approach infinity in (4) gives

soo 0= 10 + S0 - S 00 + plny.

0

If / 0 is very small compared to S0, then we can neglect it, and write

s<YJ 0= S0 - S<YJ +plnSo

Now, if S0 - pissmall compared top, then S0 - S<YJ will be small compared to S0. Consequently, we can truncate the Taylor series

[ _ ( So- S <YJ ) l = _ ( S0 - S <YJ ) _ _!_ ( S0 - S 00 )2

In 1 S S 2 S + ... 0 0 0

after two terms. Then,

= (S0 - S<YJ )[ 1- ;0

- 2; 5 (So- Soo) J. 461


Solving for S0 - S 00 , we see that

So-Soo=2S0 ( : 0 -1)=2(p+v)[ p;v -1]

=2(p+ v)~ =2p(l + ~) ~ ;;;;2v. p p p 0

During the course of an epidemic it is impossible to accurately ascertain the number of new infectives each day or week, since the only infectives who can be recognized and removed from circulation are those who seek medical aid. Public health statistics thus record only the number of new removals each day or week, not the number of new infectives. Therefore, in order to compare the results predicted by our model with data from actual epidemics, we must find the quantity dR/ dt as a function of time. This is accomplished in the following manner. Observe first that

Second, observe that

dR dt = y/= y(N- R- S).

dS dSjdt -rSI -S dR = dRjdt = -y! = -p-

Hence, S(R)=S0e-R/p and

~~ =y(N-R-S0e-RIP). (5)

Equation (5) is separable, but cannot be solved explicitly. However, if the epidemic is not very large, then R j p is small and we can truncate the Taylor series

2

e-R/p=l-~+t(~) + ...

after three terms. With this approximation,

~~ =y[N-R-s0 [ 1-Rjp+t(Rjp)2 ]]

The solution of this equation is

(6)

462


where

1 1 (So ) <j>=tanh- ~ P-I

and the hyperbolic tangent function tanhz is defined by ez-e-z

tanhz= z z. e +e

lt is easily verified that

Hence,

~tanhz=sech2z= 4 . dz (ez+e-z)2

dR ya_2p2 2( 1 ) dt = 2S0

sech 2ayt-<j> . (7)

Equation (7) defines a symmetric bell shaped curve in the t-dR/ dt plane (see Figure 2). This curve is called the epidemic curve of the disease. lt illustrates very weil the common observation that in many actual epidemics, the number of new cases reported each day climbs to a peak value and then dies away again.

dR dt

2</>ltXY Figure 2

Kermack and McKendrick compared the values predicted for dR/ dt from (7) with data from an actual plague in Bombay which spanned the last half of 1905 and the firsthalf of 1906. They set

dR dt = 890sech2(0.2t- 3.4)

with t measured in weeks, and compared these values with the number of deaths per week from the plague. This quantity is a very good approximation of dR/ dt, since almost all cases terminated fatally. As can be seen from Figure 3, there is excellent agreement between the actual values of

463


200

5 10 15 20 25 30

WEEKS

Figure 3

dR/ dt, denoted by •, and the values predicted by (7). This indicates, of course, that the system of differential equations (I) is an accurate and reliable model of the spread of an infectious disease within a population of fixed size.

References Bailey, N. T. J., 'The mathematical theory of epidemics,' 1957, New York.

Kermack, W. 0. and McKendrick, A. G., Contributions to the mathematical the-ory of epidemics, Proceedings Roy. Stat. Soc., A, 115, 700-721, 1927.

Waltman, P., 'Deterministic threshold models in the theory of epidemics,' Springer-Verlag, New York, 1974.

EXERCISES

1. Derive Equation (6).

2. Suppose that the members of (S) are vaccinated agairrst the disease at a rate ll. proportional to their number. Then,

dS -= -rSI-li.S dt '

(a) Find the orbits of (*).

dl - =rSI-yl dt . (*)

(b) Conclude from (a) that S (t) approaches zero as t approaches infinity, for every solution S ( t), I ( t) of (*).

464

4.13 A model for the spread of gonorrhea

3. Suppose that the members of (S) are vaccinated against the disease at a rate A. proportional to the product of their numbers and the square of the members of (/). Then,

dS , 2 - = -rSI-1\SI dt '

(a) Find the orbits of (*).

di - =I(rS-y) dt .

(b) Will any susceptibles remain after the disease dies out?

(•)

4. The intensity i of an epidemic is the proportion of the total number of susceptibles that finally contracts the disease. Show that

where S 00 is a root of the equation

S = Soe<S- So-10 )/P,

5. Compute the intensity of the epidemic if p= 1000, I 0 = 10, and (a) S0 = 1100, (b) S0 = 1200, (c) S0 = 1300, (d) S0 = 1500, (e) S0 = 1800, (f) S0 = 1900. (This cannot be done analytically.)

6. Let R 00 denote the total number of individuals who contract the disease. (a) Show that R 00 = I0 + S0 - S 00 •

(b) Let R1 denote the members of (R) who are removed from the population prior to the peak of the epidemic. Compute R 11 R 00 for each of the values of S0 in 5a-5f. Notice that most of the removals occur after the peak. This type of asymmetry is often found in actual notifications of infectious diseases.

7. It was observed in London during the early 1900's, that large outbreaks of measles epidemics recurred about once every two years. The mathematical biologist H. E. Soper tried to explain this phenomenon by assuming that the stock of susceptibles is constantly replenished by new recruits to the population. Thus, he assumed that

dS -= -rSI+n dt r•

for some positive constants r, y, and p..

di - =rSI-yi dt

(a) Show that S = y Ir, I= !LI y is the only equilibrium solution of (*).

(*)

(b) Show that every solution S(t), I(t) of (*) which starts sufficiently close to this equilibrium point must ultimately approach it as t approaches infinity.

(c) lt can be shown that every solution S(t), I(t) of (*) approaches the equilibrium solution S = y Ir, I= !LI y as t approaches infinity. Conclude, therefore, that the system (*) does not predict recurrent outbreaks of measles epidemics. Rather, it predicts that the disease will ultimately approach a steady state.


Gonorrhea ranks first today among reportable communicable diseases in the United States. There are more reported cases of gonorrhea every year than the combined totals for syphilis, measles, mumps, and infectious hepatitis. Public health officials estimate that more than 2,500,000 Ameri-

465


cans contract gonorrhea every year. This painful and dangerous disease, which is caused by the gonococcus germ, is spread from person to person by sexual contact. A few days after the infection there is usually itching and burning of the genital area, particularly while urinating. About the same time a discharge develops which males will notice, but which females may not notice. Infected women may have no easily recognizable symptoms, even while the disease does substantial internal damage. Gonorrhea can only be cured by antibiotics (usually penicillin). However, treatment must be given early if the disease is to be stopped from doing serious darnage to the body. If untreated, gonorrhea can result in blindness, sterility, arthritis, heart failure, and ultimately, death.

In this section we construct a mathematical model of the spread of gonorrhea. Our work is greatly simplified by the fact that the incubation period of gonorrhea is very short (3-7 days) compared to the often quite long period of active infectiousness. Thus, we will assume in our model that an individual becomes infective immediately after contracting gonorrhea. In addition, gonorrhea does not confer even partial immunity to those individuals who have recovered from it. Immediately after recovery, an individual is again susceptible. Thus, we can split the sexually active and promiscuous portion of the population into two groups, susceptibles and infectives. Let c1(t) be the total number of prorniscuous males, cit) the total number of promiscuous females, x(t) the total number of infective males, and y(t) the total number of infective females, at time t. Then, the total numbers of susceptible males and susceptible females are c1(t)x(t) and cit)-y(t) respectively. The spread of gonorrhea is presumed to be governed by the following rules:

1. Male infectives are cured at a rate a1 proportional to their total number, and female infectives are cure.d at a rate a2 proportional to their total number. The constant a1 is larger than a2 since infective males quickly develop painful symptoms and therefore seek prompt medical attention. Fernale infectives, on the other hand, are usually asymptomatic, and therefore are infectious for much Ionger periods.

2. New infectives are added to the male population at a rate b1 proportional to the total number of male susceptibles and female infectives. Sirnilarly, new infectives are added to the female population at a rate b2 proportional to the total number of female susceptibles and male infectives.

3. The total numbers of prorniscuous males and prorniscuous females remain at constant levels c1 and c2, respectively.

It follows immediately from rules 1-3 that

dx dt =-a1x+b1(c 1-x)y

(1)

466


Remark. The system of equations (I) treats only those cases of gonorrhea which arise from heterosexual contacts; the case of homosexual contacts (assuming no interaction between heterosexuals and homosexuals) is treated in Exercises 5 and 6. The number of cases of gonorrhea which arise from homosexual encounters is a small percentage of the total number of incidents of gonorrhea. Interestingly enough, this situation is completely reversed in the case of syphilis. Indeed, more than 90% of all cases of syphilis reported in the state of Rhode Island during 1973 resulted from homosexual encounters. (This statistic is not as startling as it first appears. Within ten to ninety days after being infected with syphilis, an individual usually develops a chancre sore at the spot where the germs entered the body. A homosexual who contracts syphilis as a result of anal intercourse with an infective will develop a chancre sore on his rectum. This individual, naturally, will be reluctant to seek medical attention, since he will then have to reveal his identity as a homosexuaL Moreover, he feels no sense of urgency, since the chancre sore is usually painless and disappears after several days. With gonorrhea, on the other hand, the symptoms are so painful and unmistakable that a homosexual will seek prompt medical attention. Moreover, he need not reveal his identity as a homosexual since the symptoms of gonorrhea appear in the genital area.)

Our first step in analyzing the system of differential equations (1) is to show that they are realistic. Specifically, we must show that x(t) and y(t) can never become negative, and can never exceed c1 and c2, respectively. This is the content of Lemmas 1 and 2.

Lemma 1. If x(t0) and y(t0) are positive, then x(t) and y(t) arepositive for a/1 t;;.. t0•

Lemma 2. If x(t0) is less than c1 and y(t0) is less than c2, then x(t) is less than c1 andy(t) is less than c2 for all t;;.. t0•

PROOF OF LEMMA 1. Suppose that Lemma 1 is false. Let t* > t0 be the first time at which either x or y is zero. Assurne that x is zero first. Then, evaluating the first equation of (1) at t=t* gives x(t*)=b1c1y(t*). This quantity is positive. (Note that y ( t*) cannot equal zero since x = 0, y = 0 is an equilibrium solution of (1).) Hence, x(t) is less than zero for t close to, and less than t*. But this contradicts our assumption that t* is the first time at which x(t) equals zero. We run into the same contradiction if y(t*) =0. Thus, both x(t) andy(t) arepositivefort;;.. t0• D PRooF OF LEMMA 2. Suppose that Lemma 2 is false. Let t* > t0 be the first time at which either x=c1, or y=c2• Suppose that x(t*)=c1• Evaluating the first equation of (1) at t=t* gives x(t*)= -a1c1• This quantity is negative. Hence, x(t) is greater than c1 fort close to, and less than t*. Butthis

467


contradicts our assumption that t* is the firsttime at which x(t) equals c1•

We run into the same contradiction if y(t*) = c2• Thus, x(t) is less than c1

andy(t) is less than c2 fort> t0• D

Having shown that the system of equations (1) is a realistic model of gonorrhea, we now see what predictions it makes concerning the future course of this disease. Will gonorrhea continue to spread rapidly and uncontrollably as the data in Figure 1 seems to suggest, or will it Ievel off eventually? The following extremely important theorem of epidemiology provides the answer to this question.

Theorem 8. (a) Suppose that a1a2 is less than b1b2c1c2• Then, every solution x(t),

y(t) of (1) with 0 < x(t0) < c1 and 0 < y(t0) < c2, approaches the equilibrium solution

as t approaches infinity. In other words, the total numbers of infective males and infective females will ultimately Ievel off.

(b)Suppose that a1a2 is greater than b1b2c1c2• Then every solution x(t), y(t) of (1) with 0< x(t0) < c1 and O<y(t0) < c2, approaches zero as t approaches infinity. In other words, gonorrhea will ultimately die out.

Our first step in proving part (a) of Theorem 8 is to split the reetangle 0 < x < c1, 0 <y < c2 into regions in which both dx I dt and dy I dt have fixed signs. This is accomplished in the following manner. Setting dx I dt = 0 in (1), and solving for y as a function of x gives

Similarly, setting dy I dt = 0 in ( 1) gives

Observe first that <P1(x) and <Pz(x) are monotonic increasing functions of x; <P1(x) approaches infinity as x approaches c1, and <Pz(x) approaches c2 as x approaches infinity. Second, observe that the curvesy=<P1(x) andy=<Pz(x) intersectat (0,0) and at (x0, y 0) where

468

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55

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50

0

45

0·-

40

0

35

0

30

0

25

0

20

0·-

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I f

19

50

1

95

2

19

54

19

56

19

58

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96

0

19

62

1

96

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70

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97

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y=~(x) I

y

Cz~-------------------------7~-------,

x> 0 y< 0

Figure 2

Third, observe that cJ>ix) is increasing faster than cp1(x) at x=O, since

Hence, cp2(x) lies above cp1(x) for 0<x<x0, and cJ>ix) lies below cp 1(x) for x0 < x < c1, as shown in Figure 2. The point (x0, y0) is an equilibrium point of (I) since both dx I dt and dy I dt are zero when x = x0 and y = y 0•

Finally, observe that dxl dt is positive at any point (x,y) above the curvey=cJ>1(x), and negative at any point (x,y) below this curve. Similarly, dyldt is positive at any point (x,y) below the curvey=cp2(x), and negative at any point (x,y) above this curve. Thus, the curves y =cp1(x) and y = cp2(x) split the reetangle 0 < x < c1, 0 <y < c2 into four regions in which dx I dt and dy I dt have fixed signs (see Figure 2).

Next, we require the following four simple Iemmas.

Lemma 3. Any solution x(t), y(t) of (I) which starts in region I at timet= t0

will remain in this region for all future time t ;;. t0 and approach the equilibrium solution x = x0, y = y 0 as t approaches infinity.

PROOF. Suppose that a solution x(t),y(t) of (I) leaves region I at timet= t*. Then, either x(t*) or y(t*) is zero, since the only way a solution of (I) can leave region I is by crossing the curve y =cJ>1(x) or y =cp2(x). Assurne that .X(t*)=O. Differentiating both sides of the first equation of (I) with re-

470


spect to t and setting t = t* gives

This quantity is positive, since x(t*) is less than c1, and dy I dt is positive on the curvey=cp1(x), 0<x<x0• Hence, x(t) has a minimum at t=t*. But this is impossible, since x(t) is increasing whenever the solution x(t), y(t) is in region I. Similarly, if y(t*)=O, then

This quantity is positive, since y( t*) is less than c2, and dx I dt is positive on the curvey=cpix), 0<x<x0• Hence,y(t) has a minimum at t=t*. But this is impossible, since y(t) is increasing whenever the solution x(t), y(t) is in region I.

The previous argument shows that any solution x(t),y(t) of (I) which starts in region I at time t = t0 will remain in region I for all future time t;.. t0• This implies that x(t) and y(t) are monotonic increasing functions of time fort;.. t0 , with x(t)<x0 andy(t)<y0• Consequently, by Lemma I of Section 4.8, both x( t) and y ( t) have Iimits ~. 1J, respectively, as t approaches infinity. Lemma 2 of Section 4.8 implies that (~. 1J) is an equilibrium point of (1). Now, it is easily seen from Figure 2 that the only equilibrium points of (1) are (0,0) and (x0, y 0). But (~,7J) cannot equal (0,0) since both x(t) and y(t) are increasing functions of time. Hence, (~;1J)=(x0, y 0), and this proves Lemma 3. D

Lemma 4. Any solution x(t), y(t) of (1) which starts in region Ill at timet= t0 will remain in this region for a/1 future time and ultimately approach the equilibrium solution x = x 0, y = Yo·

PROOF. Exactly the same as Lemma 3 (see Exercise 1). D

Lemma 5. Any solution x(t), y(t) of (1) which starts in region II at time t = t0, and remains in region II for a/1 future time, must approach the equilibrium solution x = x0, y = y 0 as t approaches infinity.

PROOF. If a solution x(t), y(t) of (1) remains in region II for t;.. t0 , then x(t) is monotonic decreasing and y(t) is monotonic increasing for t;.. t0•

Moreover, x(t) is positive andy(t) is less than c2, fort;.. t0• Consequently, by Lemma 1 of Section 4.8, both x(t) andy(t) have Iimits ~.1/ respectively, as t approaches infinity. Lemma 2 of Section 4.8 implies that (~, 1J) is an equilibrium point of (1). Now, (~,7J) cannot equal (0,0) sincey(t) is increasing fort;.. t0 • Therefore, (~,1J)=(x0,y0), and this proves Lemma 5. D

471


Lemma 6. Any solution x(t), y(t) of (I) which starts in region IV at timet= t0 and remains in region IV for al/ future time, must approach the equilibrium solution x = x0, y = y 0 as t approaches infinity.

PR.ooF. Exactly the same as Lemma 5 (see Exercise 2). D

W e are now in a position to prove Theorem 8.

PROOF OF THEOREM 8. (a) Lemmas 3 and 4 state that every solution x(t), y(t) of (1) which starts in region I or III at time t = t0 must approach the equilibrium solution x = x 0, y = y 0 as t approaches infinity. Similarly, Lemmas 5 and 6 state that every solution x(t), y(t) of (I) which starts in region li or IV and which remains in these regions for all future time, must also approach the equilibrium solution x = x0, y = y 0• Now, observe that if a solution x(t), y(t) of (I) leaves region li or IV, then it must cross the curve y =<t>1(x) or y =<t>2(x), and immediately afterwards enter region I or region III. Consequently, all solutions x(t), y(t) of (I) which start in regions II and IV or on the curves y =cf>1(x) and y =<t>2(x), must also approach the equilibrium solution x(t) = x0,y(t) = y 0• D

(b) PROOF #I. If a1a2 is greater than b1b2c1c2, then the curves y =<t>1(x) and y =<f>lx) have the form described in Figure 3 below. In region I, dx I dt is positive and dy I dt is negative; in region II, both dx I dt and dy I dt are negative; and in region III, dx I dt is negative and dy I dt is positive. lt is a simple matter to show (see Exercise 3) that every solution x(t), y(t) of (I) which starts in region II at timet= t0 must remain in this region for all

Figure 3

472


future time, and approach the equilibrium solution x = 0, y = 0 as t approaches infinity. 1t is also trivial to show that every solution x(t), y(t) of (1) which starts in region I or region 111 at time t = t0 must cross the curve y=cPJ(x) or y=cf>ix), and immediately afterwards enter region II (see Exercise 4). Consequently, every solution x(t),y(t) of (1), with O<x(t0)<c1

and O<y(t0)< c2, approaches the equilibrium solution x=O, y =0 as t approaches infinity. D

PRooF #2. We would now like to show how we can use the PoincareBendixson theorem to give an elegant proof of part (b) of Theorem 8. Observe that the system of differential equations (1) can be written in the form

(2)

Thus, by Theorem 2 of Section 4.3, the stability of the solution x = 0, y = 0 of (2) is determined by the stability of the equilibrium solution x = 0, y = 0 of the linearized system

~(~)=A(~)=( ~;~ ~:J(~)· The characteristic polynomial of the matrix A is

A2 +(a1 + a2)A+ a1a2 - b1b2c1c2

whose roots are

[ 2 ]1/2 -(a1+a2)± (a 1+a2) -4(a1a2-b1b2c1c2) A= 2

lt is easily verified that both these roots are real and negative. Hence, the equilibrium solution x = 0, y = 0 of (2) is asymptotically stable. This implies that any solution x(t), y(t) of (1) which starts sufficiently close to the origin x = y = 0 will approach the origin as t approaches infinity. Now, suppose that a solution x(t),y(t) of (1), with 0<x(t0)<c1 and O<y(t0)< c2, does not approach the origin as t approaches infinity. By the previous remark, this solution must always remain a minimum distance from the origin. Consequently, its orbit for t ~ t0 lies in a bounded region in the xy plane which contains no equilibrium points of (1). By the PoincareBendixson Theorem, therefore, its orbit must spiral into the orbit of a periodic solution of (1). But the system of differential equations (1) has no periodic solution in the first quadrant x ~ 0, y ~ 0. This follows immediate1y from Exercise 11, Section 4.8, and the fact that

a a ax[ -alx+bl(cl-x)y]+ ay[-a2y+bic2-y)x]

= -(a1 +a2 +Ö1y+b2x)

is strictly negative if both x and y are nonnegative. Consequently, every

473


solution x(t), y(t) of (1), with 0<x(t0)<c1 and 0<y(t0)<c2 approaches the equilibrium solution x = O,y = 0 as t approaches infinity. 0

Now, it is quite difficult to evaluate the coefficients a 1, a2, b1, o2, c1, and c2• Indeed, it is impossible to obtain even a crude estimate of a2, which should be interpreted as the average amount of time that a female remains infective. (Similarly, a1 should be interpreted as the average amount of time that a male remains infective.) This is because most females do not exhibit symptoms. Thus, a female can be infective for an amount of time varying from just one day to well over a year. Nevertheless, it is still possible to ascertain from public health data that a1a2 is less than b1b2c1c2, as we now show. Observe that the condition a 1a2 < b1b2c1c2 is equivalent to

The quantity b1c1/ a2 can be interpreted as the average number of males that one female infective contacts during her infectious period, if every male is susceptible. Similarly, the quantity b2c2/ a 1 can be interpreted as the average number of females that one male infective contacts during his infectious period, if every female is susceptible. The quantities b1c1/ a2 and b2c2/ a1 are called the maximal female and male contact rates, respectively. Theorem 8 can now be interpreted in the following manner.

(a) If the product of the maximal male and female contact rates is greater than one, then gonorrhea will approach a nonzero steady state.

(b) If the product of the maximal male and female contact rates is less than one, then gonorrhea will die out eventually.

In 1973, the average number of female contacts named by a male infective during his period of infectiousness was 0.98, while the average number of male contacts named by a female infective during her period of infectiousness was 1.15. These numbers are very good approximations of the maximal male and female contact rates, respectively, and their product does not exceed the product of the maximal male and female contact rates. (The number of contacts of a male or female infective during their period of infectiousness is slightly less than the maximal male or female contact rates. However, the actual number of contacts is often greater than the number of contacts named by an infective.) The product of 1.15 with 0.98 is 1.0682. Thus, gonorrhea will ultimately approach a nonzero steady state.

Remark. Our model of gonorrhea is rather crude since it lumps all promiscuous males and all promiscuous females together, regardless of age. A more accurate model can be obtained by separating the male and female populations into different age groups and then computing the rate of change of infectives in each age group. This has been done recently, but the analysis is too difficult to present here. We just mention that a result

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completely analogous to Theorem 8 is obtained: either gonorrhea dies out in each age group, or it approaches a constant, positive Ievel in each age group.

EXERCISES

In Problems 1 and 2, we assume that a1a2 < b1b2c1c2•

1. (a) Suppose that a solution x(t), y(t) of (1) leaves region III of Figure 2 at time t = t* by crossing the curve y =ci>I(x) or y =cp2(x). Conclude that either x(t) or y(t) has· a maximum at t= t*. Then, show that this is impossible. Conclude, therefore, that any solution x(t), y(t) of (1) which starts in region III at time t = t0 must remain in region III for all future time t > t0.

(b) Conclude from (a) that any solution x(t), y(t) of (1) which starts in region III has a 1imit ~. 11 as t approaches infinity. Then, show that (~, 'IJ) must equal (xo. Yo).

2. Suppose that a solution x(t), y(t) of (1) remains in region IV of Figure 2 for all time t > t0. Prove that x(t) and y(t) have Iimits ~. 11 respectively, as t approaches infinity. Then conclude that (~, 'IJ) must equal (x0, y 0).

In Problems 3 and 4, we assume that a1a2> b1b2c1c2.

3. Suppose that a solution x(t), y(t) of (1) leaves region II of Figure 3 at time t = t* by crossing the curve y =cp1(x) or y =cp2(x). Show that either x(t) or y(t) has a maximum at t = t*. Then, show that this is impossible. Conclude, therefore, that every solution x(t),y(t) of (1) which starts in region II at timet= t0 must remain in region II for all future time t > t0 .

4. (a) Suppose that a solution x(t), y(t) of (1) remains in either region I or III of Figure 3 for all time t > t0• Show that x( t) and y ( t) have Iimits ~. 11 respectively, as t approaches infinity.

(b) Conclude from Lemma 1 of Section 4.8 that (~,1J)=(O,O). (c) Show that (~,'IJ) cannot equal (0,0) if x(t),y(t) remains in region I or region

III for all time t > t0•

(d) Show that any solution x(t),y(t) of (1) which starts on either y =cp1(x) or y = cp2(x) will imrnediately afterwards enter region II.

5. Assurne that a1a2 < b1b2c1c2• Prove directly, using Theorem 2 of Section 4.3, that the equilibrium solution x=x0,y=y0 of (1) is asymptotically stable. Warning: The calculations are extremely tedious.

6. Assurne that the number of homosexuals remains constant in time. Call this constant c. Let x(t) denote the number of homosexuals who have gonorrhea at time t. Assurne that homosexuals are cured of gonorrhea at a rate a 1, and that new infectives are added at a rate ß1(c-x)x. (a) Show that i= -a1x+ß1x(c-x). (b) What happens to x(t) as t approaches infinity?

7. Suppose that the number of homosexuals c(t) grows according to the logistic 1aw c=c(a-bc), for some positive constants a and b. Let x(t) derrote the number of homosexuals who have gonorrhea at timet, and assume (see Problem 6) that x= -alx+ßJx(c-x). What happens to x(t) as t approaches infinity?

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